Class 11 Chemistry

Question 1. Discuss whether the difference between the heat of the reaction at constant pressure and constant volume will depend on the temperature of the following reaction.
Answer: 

2CO(g) + O₂(g)→+2CO₂(g)

We know, AH = AU+AnRT

For the given reaction, An = 2-(2 + 1) = -1

ΔH-ΔU = ΔnRT = -RT

So, the value of (ΔH-ΔU) depends on the temperature.

Question 2. Calculate the change in internal energy of the gas when it expands from 2L to 8L at a constant pressure of 2 atm absorbing 400 J of heat in the process. (1 L-atm = 101.35 J)
Answer:

We know, w = -Pex (V2-V1)

∴ ω=_2(8- 2)=-12 L.atm=-12 x 101.3 J =-1215.6 J.

According to the 1st law of thermodynamics, A U=q + ω

Given that q = 400 J

∴ ΔU= (400- 1215.6)J = -815.61

∴ Change in internal energy of the gas =-815.61.

thermodynamics class11 chemistry

Question 3. At 25 °C the standard heat of formation of liquid H20 is -286.0 kJ mol-1. Calculate the change in standard internal energy for this formation reaction.
Answer:

Formation reaction of \(\mathrm{H}_2 \mathrm{O}(l): \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

For this rections \(\Delta n=0-\left(1+\frac{1}{2}\right)=-\frac{3}{2} \text {. }\)

We know, ΔH° = ΔU° + ΔnRT

∴ \(-286.0 \mathrm{~kJ}=\Delta U^0+\left(-\frac{3}{2}\right) \times 8.314 \times 10^{-3} \times(273+25) \mathrm{kJ}\)

or, ΔU° = -282.28 kj

Therefore, the change in internal energy for the formation reaction of H20(/) is -282.28 kJ.

Question 4. The temperature of 4 mol of a gas decreases from 40°C to -60 °C on adiabatic reversible expansion. The molar-specific heat of the gas at a constant volume being 12 J.K-1 mol-1, determines the change in internal energy and work done in this process
Answer:

For an adiabatic expansion

\(w=\Delta U=n C_{V, m}\left(T_2-T_1\right)\left[T_2<T_1\right]\)

∴ ΔU = 4 X 12 X (213- 313) J = -4.8 kj and w = Δt = -4.8 kj

∴ In the process, the change in internal energy = -4.8 kJ, and the amount of work done = 4.8 kJ.

Question 5. Write the thermodynamic relation generally used to predict whether a reaction is spontaneous or not For exothermic and endothermic reactions with their change in entropies being positive and independent of temperature variations, comment on the spontaneity of the reactions in both cases for temperature variations. Compute ΔH° at 298 K for: OH(g)→+H(g) + O(g)
Answer:

The thermodynamic relation that is generally used to predict the spontaneity of a reaction is AG = AH- TAS, where AG, AH, and AS are the changes in free energy, enthalpy, and entropy in the reaction at a given pressure and temperature of T K. At a given pressure and temperature, for a spontaneous reaction AG < 0. In exothermic reaction, AH < 0.

In such a reaction, if AS> 0, then AG = AH-TAS = -ve-T(+ve). As T is always positive, AG will always be negative at any value of T. Hence, an exothermic reaction will always be spontaneous at any temperature if AS > 0.

In an endothermic reaction, AH > 0. In such a reaction, if AS >0, then AG = AH-TAS =+ve-T(+ve) So, AG will be -ve only when |TASl > |AH|. It happens at high temperatures. So, an endothermic reaction with AS > 0 will be spontaneous at high temperatures.

chapter 6 chemistry class 11

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{OH}(\mathrm{g}) ; \Delta H^0=10.06 \mathrm{kcal} \quad \cdots[1]\)

⇒ \(\begin{aligned}
& \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g}) ; \Delta H^0=104.18 \mathrm{kcal} \\
& \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{O}(\mathrm{g}) ; \Delta H^0=118.32 \mathrm{kcal}
\end{aligned}\)

By dividing each of the equations [2] and [3] by 2, and then adding them together, we have

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \text {; }\)

ΔHº= (52.09 + 59.161) kcal = 111.25 kcal

Subtracting equation [1] from equation [4], we have

OH(g)→H(g) + O(g) ; ΔHº = (111.25 -10.06) kcal = 101.19 kcal

So, ΔHº for the given reaction is 101.19 kcal.

Question 6. For which of the following reactions, AS > 0 —

1. \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)
2. \(\mathrm{HCl}(g)+\mathrm{NH}_3(g) \rightleftharpoons \mathrm{NH}_4 \mathrm{Cl}(s)\)
3. \(\mathrm{NH}_4 \mathrm{NO}_3(s) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(g)\)
4. \(\mathrm{MgO}(s)+\mathrm{H}_2(g) \rightleftharpoons \mathrm{Mg}(s)+\mathrm{H}_2 \mathrm{O}(g)\)

Answer: 3. \(\mathrm{NH}_4 \mathrm{NO}_3(s) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(g)\)

⇒ \(\mathrm{NH}_4 \mathrm{NO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})\)

Question 7. The heat of combustion of benzene is x J-mol-1. The heat of the formation of carbon dioxide and water are y J.mol-1 and k J .mol-1 respectively. Calculate the heat of the formation of benzene.
Answer: \(\mathrm{C}_6 \mathrm{H}_6(l)+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(l) ;\)

With increasing the number of species in the system, the entropy system increases for which AS > 0.

ΔH = -x J-mol-1

C(s) +O₂(g) CO₂(g) ; AH = -y J. mol-1

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \Delta H=-z \mathrm{~J} \cdot \mathrm{mol}^{-1}\)

6 x eqn.(2) + 3 x eqn.(3)- eqn.(1) we get,

6C(s) + 3H2(g)→CgH6(l) ;

= (x- 6y- 3z) J .mol-1

ch 6 chemistry class 11

Question 8…… The combustion reaction for methane is—

⇒ \(\mathrm{CH}_4(g)+2 \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l)\)

For this reaction, An = 1- (1 + 2) = -2

we know ΔH0 = ΔU° + ΔnRT

For the above combustion reaction, ΔH° = – X- 2RT

As T is positive, ΔH0 < ΔU0

Question 9. The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and A U was found to be -742.7 kj.mol-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K:

⇒ \(\mathrm{NH}_2 \mathrm{CN}(s)+\frac{3}{2} \mathrm{O}_2(g) \rightarrow \mathrm{N}_2(g)+\mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l)\)

thermodynamics exercises

Answer: For the given reaction. Δn \(=(1+1)-\frac{3}{2}=+\frac{1}{2}\)

We know, ΔH = ΔU+ΔnRT

∴ \(\Delta H=\left(-742.7+\frac{1}{2} \times 8.314 \times 10^{-3} \times 298\right)\)

= -741.64kJ . mol-1

∴ The change in enthalpy for the reaction is -741.46kJ.mol-1

Question 10. Calculate the number of kj necessary to raise the temperature of 60g of aluminum from 35°C to 55°C. The molar heat capacity of A1 is 24J.mol-1. K-1.
Answer: Heat (q) required to raise temp, of m g of a substance (specific heat capacity=c) from T1 to T2 is: q=mc(T2-T1)

⇒ \(c=\frac{\text { Molar heat capacity }}{\text { Molar mass }}=\frac{24 \mathrm{~J} \cdot \mathrm{K}^{-1}}{27 \mathrm{~g}}=\frac{24}{27}\left(\mathrm{~J} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\right)\)

⇒ \(\text { So, } q=60 \times \frac{24}{27}[(273+55)-(273+35)] \mathrm{J}=1.066 \mathrm{~kJ}\)

Hence, 1.066kJ heat is required.

Question 11. Calculate enthalpy change on freezing 1 mol water at 10°C to ice at -10°C. A = 6.03kJ. mol1 at 0°C. Cp[H20(J)]=75.3J.mol-1.K-1 , Cp[H20(s)]=36.8J. mol-1.K-1
Answer: The given process can be considered as the sum of the following two processes—

Water(10°C) -4 water(0°C) 4 ice(0°C) 4 ice (-10°C)

To calculate enthalpy changes in steps (1) and (3), we use the relation, qp=AH=Cp(T2-T1) [At constant P, q = AH].

Step 1: AHj =Cp[H2O(l)](T2-T1)

= 75.31. mol-1.K-1 [273- (273 + 10)]K

=-0.753 kl-mol-1

Step 2: Freezing of water takes place. In this step,

AH2 = -AfusH = -6.03kj.mol-1

Step 3: AH3 =Cp[H20(s)](T2- TJ)

= 36.8[(273—10)—(273+0)] =-0.368 kj-mol-1

Step 2: Freezing of water takes place. In this step

ΔH2 = -ΔfusH = -6.03kj. mol-1

Step 3: AH3 =Cp[H20(s)](T2- TJ)

= 36.8[(273—10)—(273+0)] =-0.368 kl-mol-1

So, the total change in enthalpy in the entire process,
AH =AHJ+AH2 + AH3

=[- 0.753- 6.03- 0.368]kJ. mol-1

=-7.151kj.mol-1

Note: The answer to this problem is given as -5.65kJ.mol-1. The process may be: water at -10°C -> ice at -10°C,

solutions thermodynamics

Question 12. Enthalpy of combustion of C to CO₂: -393.5 kj.mol-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Answer:

Given:C(s) + O₂(g)→CO₂(g); AH0 = -393.5 kj.mol-1

According to this equation, 44 g of CO₂ = 12 g of C

∴ \(35.2 \mathrm{~g} \text { of } \mathrm{CO}_2 \equiv \frac{12}{44} \times 35.2 \equiv 9.6 \mathrm{~g} \text { of } \mathrm{C} \equiv 0.8 \mathrm{~mol} \text { of } \mathrm{C}\)

From above equation, 1 mol C = 393.5 kJ of heat released

∴ 0.8 mol of C = 393.5 x 0.8kJ = 314.8kJ of heat released Therefore, the heat released on the formation of 35.2 g of CO₂ from C and O₂ is 314.8 kj

Question 13. Enthalpies of formation of CO(g), CO₂(g) , N₂O(g) and N₂O₄(g) are -110, -393, 81 and 9.7kj-mol-1 respectively. Find the value of ArH for the reaction: N₂O₄(g) + 3CO(g)→N₂O(g) + 3CO₂(g)
Answer:

⇒ \(\begin{aligned}
& \Delta_r H=\Sigma \Delta_f H \text { (Products) }-\Sigma \Delta_f H \text { (Reactants) } \\
&=3 \Delta_f H\left[\mathrm{CO}_2(g)\right]+\Delta_f H\left[\mathrm{~N}_2 \mathrm{O}(g)\right] \\
&-\Delta_f H\left[\mathrm{~N}_2 \mathrm{O}_4(g)\right]+3 \Delta_f H[\mathrm{CO}(g)] \\
&= {[3(-393)+1(81)-1(9.7)-3(-110)]=-777.77 \mathrm{~kJ} }
\end{aligned}\)

Question 14. Calculate the standard enthalpy of the formation of CH3OH(l) from the following data:

⇒ \(\begin{aligned}
\mathrm{CH}_3 \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta_r \mathrm{H}^0=-726 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

⇒ \(\begin{aligned}
& \mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta_c H^0=-393 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta_f H^0=-286 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)
Answer: To get the formation reaction for CH3OH(l), we combine three thermochemical equations provided in the given way:

\(\begin{array}{r}
\left.\mathrm{C}(s)+\mathrm{O}_2(\mathrm{~g})+2\left[\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\right]-\mathrm{CH}_3 \mathrm{OH}(l)-\frac{3}{2} \mathrm{O}_2(\mathrm{~g})\right\rceil \\
\mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)-\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{H}_2 \mathrm{O}(l) ;
\end{array}\)

AH0 =[- 393 + 2(—286)- (-726)]kJ.mol-1 or, C(s) + 2H₂(g) + O₂(g)→CH₃OH(Z); AH0 = -239 kj-mol-1 Eq. [1] represents formation reaction of CH3OH(Z). So, standard enthalpy offormation is -239 kl.mol-1.

Question 15. Calculate the enthalpy change for the process CCl₄(g)→C(g) + 4Cl(g) & calculate the bond enthalpy of C—Cl in CCl₄(g). Given: AvapH°(CCl4) = 30.5 kj-mol1, Af H°(CCl4) = -135.5 kl.mol-1, AaH°(C)=715.0 kj. mol-1, AaH°(Cl₂) =242 kj-mol-1 where AaH° is enthalpy of atomisation. m For the reaction 2Cl(g)→Cl₂(g) , what are the signs of AH and AS?
Answer:

Based on the given information, we can write the following thermochemical equations-

⇒ \(\begin{aligned}
& \mathrm{CCl}_4(l) \rightarrow \mathrm{CCl}_4(g) ; \Delta H^0=30.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \\
& \mathrm{C}(s)+2 \mathrm{Cl}_2(g) \rightarrow \mathrm{CCl}_4(l) ; \Delta H^0=-135.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

C(s)-C(g) ; ΔH0 = 715.0 kl. mol-1

Cl2(g)→2Cl(g) ; ΔH0 = 242 kl .mol-1

By, eq. [3]- eq. [2]- eq. [1] + 2 x eq. [4] we have,

⇒ \(\begin{gathered}
\mathrm{C}(s)-\mathrm{C}(s)-2 \mathrm{Cl}_2(g)-\mathrm{CCl}_4(l)+2 \mathrm{Cl}_2(g) \\
\mathrm{C}(g)-\mathrm{CCl}_4(l)-\mathrm{CCl}_4(g)+4 \mathrm{Cl}(g)
\end{gathered}\)

ΔH° = [715.0- (-135.5)- 30.5 + 2 x 242] kj

Or, CCl4(g)C(g) + 4Cl(g); ΔH° = 1304 k

In the CC14 molecule, there are four C—Cl bonds. To break all these bonds, 1304 kj of energy is required. Hence, C—Cl bond enthalpy in CCl4(g) = 1/4x 1304 = 326 kj.mol-1.

Hydrogen Question and Answers

Question 1. Explain why concentrated HCl is not used in the laboratory preparation of H₂ gas.
Answer:

Concentrated HCl is not used for the laboratory preparation of dihydrogen because HCl, being highly volatile, gets mixed with dihydrogen.

Question 2. Write down the name and formula of a compound that on electrolysis produces dihydrogen at the anode.
Answer:

Sodium hydride(NaH).

thermodynamics class11 chemistry

Question 3. What is syngas?
Answer:

syngas

All mixtures of CO and H₂ irrespective of their composition are called synthesis gas or syngas.

Question 4. Which isotope of hydrogen is used as a tracer in organic reactions?
Answer:

Deuterium D is usually used as a tracer in determining the mechanism of organic reactions.

Question 5. Explain why dihydrogen is not suitable for balloons.
Answer:

As dihydrogen, the lightest substance known, is a highly inflammable gas, it is not suitable for balloons.

Question 6. Which bond between two atoms has the highest bond dissociation enthalpy?
Answer:

Bond dissociation enthalpy of the H —H bond is highest.

Question 7. Explain why H2 is more reactive than D2
Answer:

This is because the H —H bond dissociation enthalpy is less than the D —D bond dissociation enthalpy.

Question 8. What change is expected to take place when vegetable oils are hydrogenated?
Answer:

Carbon-carbon double bonds are converted to carbon-carbon single bonds.

Question 9. Which isotope of hydrogen is used in nuclear reactors?
Answer:

Deuterium(H or D).

Question 10. Why are ionic hydrides used as solid fuels?
Answer:

When heated, ionic hydrides decompose to evolve dihydrogen gas which ignites readily.

chapter 6 chemistry class 11

Question 11. The densities of ionic hydrides are greater than that of the metal from which they are formed—why?
Answer:

The densities of ionic hydrides are greater than that of the metal from which they are formed

This is because hydride ions (H) occupy the holes in the lattice of the metal without distorting the metal lattice.

Question 12. Give examples of two interstitial hydrides.
Answer:

CuH and FeH.

Question 13. Which gaseous compound on treatment with dihydrogen produces methanol?
Answer:

Carbon monoxide(CO).

Question 14. Give the chemical reaction that occurs when hydrogen is used as a rocket fuel.
Answer:

\(\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(l)+286 \mathrm{~kJ}\)

hydrogen solutions

Question 15. A sample of water containing KCl does not behave as hard water, but a sample of water containing CaCI₂ or MgCl₂ behaves as hard water—why?
Answer:

A sample of water containing KCl does not behave as hard water, but a sample of water containing CaCI₂ or MgCl₂ behaves as hard water

The potassium salt of soap is soluble in water and forms a lather while calcium or magnesium salt of soap is insoluble in water and does not form a lather.

Question 16. What is EDTA, a compound used to determine the hardness of water?
Answer:

EDTA

EDTA is the disodium salt of ethylenediamine tetraacetic acid \(\left[\mathrm{NaO}_2 \mathrm{C}(\mathrm{COOH}) \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{~N}(\mathrm{COOH}) \mathrm{COONa}\right]\)

 

ch 6 chemistry class 11

Question 17. Can distilled water be called deionized water?
Answer:

Distilled water can be called deionized water because it does not contain any cations and anions.

Question 18. What is the difference between the water softened by the permit process and the water softened by the organic ion exchangers?
Answer:

The difference between the water softened by the permit process and the water softened by the organic ion exchangers

Although the water softened by the permit process contains no cation, it contains various anions (for example Cl-, SO|- etc.). However, the water softened by the organic ion exchangers contains no cations and anions.

chapter 9 Chemistry, class  11 

Question 19. What will be the hardness of a sample of water, 106 g of which contains| mol A12(SO₄)₃?
Answer:

50 ppm.

Question 20. What is Calgon?
Answer:

Calgon

Sodium hexametaphosphate, Na₂[Na₄(PO₃)₆]

Question 21. Give the chemical formula of permtit.
Answer:

Na₂Al₂Si₂O₈ xH₈O.

Question 22. What is the main source of heavy water?
Answer:

The main source of heavy water

Ordinary water is the main source of heavy water.

Question 23. Can sea animals survive in distilled water?
Answer:

Sea animals cannot survive in distilled water because distilled water contains no salt or dissolved oxygen.

Question 24. Although D20 resembles H20 chemically, it is a toxic substance—why?
Answer:

D₂O is toxic because D⁺ ions react at a much slower rate than H⁺ ions in enzyme-catalyzed reactions.

Question 25. Which compound is used to color hair golden?
Answer:

Dilute solution of H₂O₂.

Question 26. What is the trade name of hydrogen peroxide used as an antiseptic?
Answer:

Perhydrol.

Question 27. What is the strength in the normality of an ‘11.2 volume’ H₂O₂ solution?
Answer:

2(N) H₂O₂ solution.

Question 28. Name a compound that suppresses the decomposition of H₂O₂.
Answer:

Acetanilide (PhNHCOCH₃).

Question 29. H₂O₂ molecule has an open-book-like structure. What is the angle between the two pages of the book in the gas phase?
Answer:

111.5°.

Question 30. Name an organic compound without peroxo bond which is used to manufacture H₂O₂.
Answer:

2-ethylanthraquinol.

Question 31. Why do most of the reactions of H2 occur at much higher temperatures?
Answer:

Due to the much higher bond dissociation enthalpy of the H —H bond, dihydrogen is quite stable and relatively inert at temperature. It dissociates into atoms at about 5000K. For this reason, most of the reactions of dihydrogen occur at much higher temperatures.

ch 6 chemistry class 11

Question 32. What characteristics do you expect from electron-deficient hydrides with respect to their structure and chemical reactivity?
Answer:

Electron-deficient hydrides have less number of electrons in the valence shell of the central atom and so, their mononuclear units do not satisfy the usual Lewis octet rule. Due to a deficiency of electrons, these hydrides act as Lewis acids and form complex entities with Lewis bases such as NH₃, H- ion, etc.

⇒ \(\mathrm{B}_2 \mathrm{H}_6+2 \ddot{\mathrm{N}}_3 \rightarrow\left[\mathrm{BH}_2\left(\mathrm{NH}_3\right)_2\right]^{+}\left[\mathrm{BH}_4\right]^{-}\)

⇒ \(\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{LiH} \rightarrow 2 \mathrm{Li}^{+}\left[\mathrm{BH}_4\right]^{-} \text {(Lithium borohydride) }\)

Question 33. Explain whyit is harmful to bathe in heavy water and use it for drinking purposes.
Answer:

Being a very hygroscopic substance, heavy water (D₂O) absorbs water from the body and thereby damages body cells. Also, it retards some cellular processes such as mitosis, cell division, and various enzyme-catalyzed reactions. For these reasons, it is harmful to bathe in heavy water and use it for drinking purposes.

Question 34. Explain why the thermal stability of H₂O₂ is very low.
Answer:

The bond dissociation enthalpy of the O —O bond presenting H₂O₂ molecule is very low (35kcal . mol^-1) i.e., the bond is very weak. For this reason, the thermal stability of H₂O₂ is extremely low.

Question 35. How the presence of H- ions be confirmed in ionic hydrides?
Answer:

In the molten state, ionic hydrides conduct electricity with the liberation of dihydrogen at the anode. This confirms the presence of hydride (H^– ) ions in them.

Question 36. How do you separate 2 allotropic forms of hydrogen?
Answer:

Ordinary hydrogen is a mixture of 75% of ortho and 25% of para-isomer at room temperature. On passing through activated charcoal kept at 20K, the para-isomer is adsorbed leaving behind the ortho-isomer. From the charcoal surface, para-hydrogen can be released by reducing pressure.

thermodynamics exercises

Question 37. Mention the difference in chemical characteristics of the two hydrides obtained when hydrogen combines with two elements having atomic numbers 17 and 20.
Answer:

The highly electronegative Cl having atomic number 17 combines with hydrogen to form the covalent hydride H —Cl. On the other hand, the highly electropositive Ca atom having atomic number 20 combines with hydrogen to form the ionic hydride CaH₂.

Question 38. Two samples of hard water contain the same cations, Ca²⁺ & Mg²⁺. One is marked as temporary and the other as permanent. In which respect do they differ?
Answer:

The two samples of water differ with respect to the anions of calcium and magnesium salts present. the sample watermarked as temporary hard water contains bicarbonates of Ca²⁺ and Mg²⁺ ions while the sample watermarked as permanent hard water contains chlorides and sulfates of Ca²⁺ and ²⁺ ions.

Question 39. Tube-well water, if left for some time, assumes brownish turbidity—explain.
Answer:

Tube-well water, if left for some time, assumes brownish turbidity

Tube-well water sometimes contains soluble ferrous bicarbonate [Fe(HCO₃)₂]. This compound, on aerial oxidation, is converted into brown ferric hydroxide, Fe(OH)₃, which remains in water as colloidal suspension, and because of this, water assumes a brownish turbidity.

⇒ \(4 \mathrm{Fe}\left(\mathrm{HCO}_3\right)_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \rightarrow \underset{\text { (brown) }}{4 \mathrm{Fe}(\mathrm{OH})_3}+8 \mathrm{CO}_2\)

Question 40. Write the reactions for:

1. Preparation of H₂O₂ from two sodium salts and
2. Preparation of D₂O₂ from potassium persulphate.

Answer:

⇒ \(\begin{aligned} & \mathrm{K}_2 \mathrm{~S}_2 \mathrm{O}_8+2 \mathrm{D}_2 \mathrm{O} \rightarrow \mathrm{D}_2 \mathrm{O}_2+2 \mathrm{KDSO}_4 \\ & \text { Potassium } \\ & \text { persulphate } \end{aligned}\)

Question 41. Between deionised water and distilled water which one is more pure and why?
Answer:

Both deionized and distilled water are free from ions. Yet distilled water is superior to deionized water in terms of purity. This is because of the fact that deionized water contains a small amount of dissolved silica and CO₂ along with some germs and organic impurities. However, distilled water prepared in a glass apparatus does not contain any impurities other than trace amounts of silica and CO₂.

Question 42. Why is Na₂O₂ used for purifying air in submarines and in crowded places?
Answer:

Na₂O₂ reacts with CO₂ of air to evolve O₂ (2Na₂O₂ + 2CO₂ —> 2Na₂CO₃ + O₂t). For this reason, Na₂O₂ is used for the purification of air in submarines and in crowded places.

solutions thermodynamics

43. Comment on the reactions of dihydrogen with

1. Chlorine
2. Sodium And
3. Copper
4. Oxide.

Answer:

1. Dihydrogen reduces chlorine (Cl) to chloride ion (Cl-) and itself gets oxidized to form H⁺ ions. These two ions (H⁺ and Cl-) share an electron pair between themselves to form a covalent molecule of hydrogen chloride (HCl)
   ⇒ \(\mathrm{H}_2(g)+\mathrm{Cl}_2(g)\rightarrow 2 \mathrm{HCl}(g)\)
2. Sodium reduces dihydrogen to hydride ion (H~) and itself gets oxidized to sodium ion (Na⁺). In this reaction, an electron gets completely transferred from Na thereby forming ionic sodium hydride (NaH).
   \(2\mathrm{Na}(s)+\mathrm{H}_2(s)\stackrel{\Delta}{\longrightarrow} 2\mathrm{Na}^{+} \mathrm{H}^{-}(s)\)
3. Dihydrogen reduces copper(II) oxide to metallic copper while it gets oxidized to form a covalent molecule of H₂O.
   ⇒ \(\stackrel{+2}{\mathrm{CuO}}\stackrel{-2}{\mathrm{O}}(s)+\stackrel{0}{\mathrm{H}_2}(s)\stackrel{\Delta}{\longrightarrow}\stackrel{0}{\mathrm{Cu}}(s)+\stackrel{+1}{\mathrm{H}_2}\stackrel{-2}{\mathrm{O}}(l)\)

Question 45. An ionic alkali metal hydride has a covalent character to some extent and it does not react with oxygen and chlorine. This hydride is used in the synthesis of another hydride. Write the formula of the hydride and what happens when it reacts with AI₂CI₆.
Answer:

Since the alkali metal hydride possesses sufficient covalent character, the hydride is of the smallest alkali metal, Li, i.e. the hydride is LiH. As LiH is quite stable, it does not react with oxygen and chlorine. Lithium hydride reacts with AI₂CI₆ to form lithium aluminum hydride (LiAlH₄) which is extensively used as a reagent in the synthesis of different organic compounds.

⇒ \(8\mathrm{LiH}+\mathrm{Al}_2\mathrm{Cl}_6\longrightarrow2\mathrm{LiAlH}_4+6\mathrm{LiCl}\)

Question 46. Sodium reacts with dlliydrogcn to form a crystalline ionic solid. It is non-volatile and a non-conductor of electricity. It also reacts vigorously with water to liberate II2 gas. Write the formula of the Ionic solid and give a reaction between the lids solid & water. What happens when the ionic solid in its molten stater Is electrolysed?
Answer:

Sodium reacts with dihydrogen to form sodium hydride which is a crystalline ionic solid.

⇒ \(2\mathrm{Na}+\mathrm{H}_2\stackrel{\Delta}{\longrightarrow}2\mathrm{Na}^{+}\mathrm{H}^{-}\)

Sodium hydride reacts with water as follows:

⇒  \(2 \mathrm{NaH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+2 \mathrm{H}_2\)

Sodium hydride, in its solid state, does not undergo electrolysis. However, in its molten state, undergoes electrolysis. However, in its molten state, NaH undergoes electrolysis to liberate dihydrogen (H₂) at the anode and metallic sodium at the cathode.

⇒ \(\mathrm{Na}^{+} \mathrm{H}^{-}(\text {molten }) \stackrel{\text { Electrolysis }}{\longrightarrow} \underset{\text { Cathode }}{2 \mathrm{Na}(\text { molten })}+\underset{\text { anode }}{\mathrm{H}_2(g)}\)

Question 47. Why is seawater not used in boilers?
Answer:

1. Sea water is hard water. It contains Mg(HCO₃)₂ and Ca(HCO₃)2, which on boiling, form a hard heat-insulating thick layer or scale of MgCO₃ and CaCO₃ on the inner surface of the boiler. Consequently, much heat is required to raise the temperature of the boiler, and thus, fuel economy is adversely affected.
2. Again at much higher temperatures, the boiler scales and the metal of the boiler expand unequally. Due to such uneven expansion, cracks are formed on the scales. When hot water comes in contact with the hot metal surface of the boiler through these cracks, it is suddenly converted into steam. Due to the high pressure, thus developed, the boiler may burst leading to accidents. Again, MgCl₂ and MgSO₄ present in seawater undergo hydrolysis to form HCl and H₂SO₄ which degrade the metallic component of the boiler. So, seawater is not used in boilers.

Question 48. The values of melting point, enthalpy of vaporization, and viscosity of H₂O and D₂O are given below:

1. Melting point(K)     373.0             374.4
2. Enthalpy of
   vaporisation
   (kJ • mol^-1, 373K)     40.66         41.61
3. Viscosity(centipoise) 0.8903      1.107

From the given data, determine which liquid has a greater magnitude of intermolecular forces of attraction.

Answer:

The magnitude of intermolecular forces of attraction depends on the magnitudes of melting point, enthalpy of vaporization, and viscosity of liquid. As, these parameters have higher values in the case of D₂O, the magnitude of intermolecular forces of attraction is greater for D₂O than for H₂O.

Question 49. How will you prepare heavy water from ordinary water? Explain its principle.
Answer:

Heavy water (D₂O) is prepared by prolonged electrolysis of ordinary water. As water is not a good conductor of electricity, an alkaline solution of water [~0.5(N) NaOH] is used for electrolysis. The bond dissociation energy of the O —H bond is less than that of the O —D bond. So, electrolysis of H₂O occurs at a faster rate and more easily than D₂O. Consequently, the amount of D₂O in ordinary water increases. Pure D₂O is obtained when the amount of residual liquid decreases.

Question 50. Can phosphorus form PHg with its outer electronic configuration of 3s²3p³?
Answer:

Phosphorus cannot form PH5 although it shows +3 and +5 oxidation states. Dihydrogen acts as a weak oxidizing agent due to the high bond dissociation enthalpy of H—Hbond (435.88 kJ-mo^-1) and slightly negative electron-gain enthalpy (-73 kJ mol^-1). So, dihydrogen can oxidize phosphorus to a +3 oxidation state but not to its highest oxidation state of +5. Therefore, phosphorous can form only PH₃ and not PH₅.

Question 51. How will you prepare dinitrogen from HNO₃?
Answer:

Magnesium and manganese react with a very dilute solution of HN03(2%) to form dihydrogen.

⇒ \(\mathrm{Mg}+2\mathrm{HNO}_3\rightarrow\mathrm{Mg}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow\)

⇒ \(\mathrm{Mn}+2\mathrm{HNO}_3\rightarrow\mathrm{Mn}\left(\mathrm{NO}_3\right)_2+\mathrm{H}_2 \uparrow\)

Question 52. Do you think the hydrides of N, O, and F will have lower boiling points than the hydrides of their corresponding group members? State reasons.
Answer:

The hydrides of the elements N, O, and F are NH₃, H₂O, and HF respectively. These hydrides are expected to have lower boiling points than that of their corresponding group members (PH₃, H₂S & HCl) when their masses are considered.
However, because of the high electronegativity of N, O, and F, their hydrides undergo extensive hydrogen bonding (intermolecular). As a result, boiling points of NH3, H₂O, and HF are much higher than the hydrides of their corresponding group members, i.e., PH₃, H₂S, and HCl respectively.

Question 53. KF reacts with HF to form the compound, KF-2HF. Discuss the probable structure of the compound.
Answer:

The h-bond in the HF molecule is very strong. When KF gets added to HF, one F^– ion forms an H — bond with two HF molecules to form the H₂F^–₃ ion [F—-H —F—H —F —].

Question 54. Calculate the amount of energy liberated due to combustion of 4g dihydrogen.
Answer:

Amount of energy liberated due to combustion of1 mol, i.e., 2g dihydrogen = 242 kJ • mol^-1

⇒ \(\left[\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{H}_2 \mathrm{O}(g) ; \Delta \mathrm{H}=-242 \mathrm{~kJ}\cdot\mathrm{mol}^{-1}\right]\)

Amount of energy liberated due to 4g dihydrogen = 242×2 = 484 kJ-mol-1

Question 55. Under what conditions, water reacts with calcium cyanamide and what are the products formed due to this reaction?
Answer:

Superheated steam reacts with calcium cyanamide under high pressure to form ammonia gas and calcium carbonate as a result of hydrolysis.
\(\text { CaNCN }(\text{Calciumcyanamide })+3\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{CaCO}_3+2\mathrm{NH}_3\)

Question 56. Why H₂O₂ is a better oxidant than water?
Answer:

H₂O₂ is a better oxidant than water because H₂O₂ being unstable readily dissociates to form stable water molecules along with the evolution of O₂ gas. \(2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2+\text { Heat }\)

Solved Questions

Question 1. What is heavy water? Why is it so-called?
Answer:

Heavy water

Deuterium oxide is commonly known as heavy water because its density is higher than that of ordinary water, i.e., it is heavier than ordinary water.

Question 2. A water sample contains 1 millimole of Mg²⁺ ion per liter. Calculate the hardness of the water sample in ppm units.
Answer:

1 millimol of Mg2+=1 millimole of MgCl2 = 0.095 g of MgCL2.

1 1. or 1000 g or 103 g of water contains 0.1 g of MgCl2.

106g of water contains 0.095 x 103 = 95g of MgCl2

The degree of hardness of water is 95 ppm.

Question 3. BaO₂ is a peroxide but MnO₂ is not a peroxide explain.
Answer:

There is peroxide linkage in the Ba02 molecule ( —O —O—). But there is no such bonding present in MnO, molecule (O —Mn=0). Thus Mn02 is not a peroxide.

 

Question 1. What will be the name (IUPAC) and symbol if the element with atomic number 119 is discovered? Write its electronic configuration. Also, write the formulas of the stable chloride and oxide of this element.
Answer:

IUPAC name: Ununennium, Symbol: Uue

• Atomic number ofthe element =119 = 87 + 32
• It is known that the element with atomic number 87 is francium (Fr). Fr belongs to group-1 in the 7th period of the periodic table.
• So, the element with atomic number 119 will take its position in group-1 and 8th period just below francium(Fr).
• The electronic configuration of this element will be [UuojBs1, (where Uuo = Ununoctium, Z = 118). It will be an alkali metal with valency=1
• If the symbol ofthe element is ‘M’ then the formulas of its stable chloride and oxide will be MCI and M20 respectively.

periodic table question answer 

Question 2. Formulas of oxide and chloride of an element A are A205 and AClg respectively. Which group of Mendeleev’s periodic table will the element belong to? State whether it is a metal or a non-metal.
Answer:

Hiking oxygen as standard, the valency ofthe element A is 5, and taking chlorine as standard, the valency of A is 3. Since oxygen-based valency (8-chlorine-based valency) is the same as that ofthe group number of the element, the element will be of group Mendeleev’s periodic table. Thus, itis a non-metal.

Question 3. A and B are two elements with atomic numbers 9 and 17 respectively. Explain why the element A is a more powerful oxidizing agent than the element B.
Answer:

Given that A and B are two elements with atomic numbers 9 and 17 respectively.

Electronic configuration of gA: ls²2s²2p⁵. Electronic configuration of B: ls²2s²2p⁶3s²3p⁵. Both A and B can accept electrons to form the anions of A¯ and B¯ having gas electronic configuration. However, the anions A¯ is more stable than B¯ because of the smaller atomic size of A. So A is a stronger oxidizing agent than B.

Question 4. Elements A, B, and C have atomic numbers (Z- 2), Z, and (Z +1) respectively. Of these, B is an inert gas. Which one of these has the highest electronegativity? Which one of these has the highest ionization potential? What is the formula of the compound formed by the combination of A and C? What is the nature of the bond in this compound?
Answer:

Since element B (atomic no =Z) is an inert gas, the element ‘A’ with atomic number (Z- 2) is included in group VIA. On the other hand, the element C, having an atomic number (Z + 1) must belong to groups (alkali metal). Hence, the electronegativity of the element A is maximum.

The element B, being an inert gas, has the highest value of ionization potential.

The valency of the element A, belonging to group (8- 6) = 2, and that ofthe element C, being an element of group IA, is 1. Therefore, the formula of the compound formed by A and C will be C2A.

A being a strongly electronegative element and C being a strongly electropositive element complete their octet through gain and loss of electrons respectively. So, the nature of the bond formed between C and A in C2A isionic or electrovalent bond.

Question 5. The atomic radius of 10Ne is more than that of gF —why?
Answer:

The atomic radius of 10Ne is more than that of gF

Fluorine forms diatomic molecules, thus the atomic radius of fluorine is a measure of half of the internuclear distance in its molecule (i.e., half of covalent bond length of F₂ molecule) but neon being an inert gas, its atoms are incapable of forming covalent bonds by mutual combination amongst themselves.

The only force that comes into play between the atoms is the weak van der Waals force. So a measure of the atomic radius of Ne is equal to its van der Waals radius but the van der Waals radius is always greater than the covalent radius.

Thus, the atomic radius of neon is larger than that of fluorine. Furthermore, due to an increase in the number of electrons in the outermost 2p -orbital of Ne, there occurs an increase in electron-electron repulsion. So 2p-orbital of Ne suffers expansion leading to its increased atomic radius.

Question 6. The first electron affinity ofoxygen is negative but the second electron affinity is positive—explain.
Answer:

The first electron affinity ofoxygen is negative but the second electron affinity is positive

When an electron is added to the valence shell of an isolated gaseous O-atom in its ground state to form a negative ion, energy is released.

Because a neutral oxygen atom tends to complete its octet with electrons. So, the electron affinity of oxygen is an exothermic process and its value is negative.

When an extra electron is added to the O¯ ion, that second electron experiences a force of repulsion exerted by the negative charge ofthe anion. So, first, this process requires a supply of energy from an external source. This accounts for the endothermic nature of second electron affinity and has a positive value.

Question 7. The electron affinity of sodium is negative but magnesium has a positive value—why?
Answer:

The electron affinity of sodium is negative but magnesium has a positive value

Electronic configuration of ₁₁Na: ls²2s²2p⁶3s¹ Electronic configuration of ₁₂Mg: ls²2s²2p⁶3s². The addition of one electron to the 3s -orbital of Na leads to a comparatively stable electronic configuration with a fulfilled orbital.

So, this process of the addition of electrons to Na is an exothermic, process. So, the electron affinity of Na is negative. On the other hand, Mg has fulfilled 3s orbital and has a stable electronic configuration. So the addition of an electron to the 3p -orbital destabilizes the electronic configuration of Mg.

Additional energy is required for addition of electrons to the outermost shell of Mg i.e., this process is endothermic and thus the value of electron affinity of Mg is positive.

Question 8. If the electron affinity of chlorine is 350 kJ.moI-1, then what is the amount of energy liberated to convert 1.775 g of chlorine (existing at atomic state) to chloride ions completely (in a gaseous state)
Answer:

Atomic mass of chlorine = 35.5 g .mol 1

The energy liberated in the conversion of 35.5 g of Cl- to Clion =350 kJ

∴ The energy liberated in the conversion of 1.775 g of Cl to Cl-

\(\text { ion }=\frac{350}{35.5} \times 1.775=17.5 \mathrm{~kJ}\)

Question 9. The second ionization enthalpy of Mg is sufficiently high and the second electron-gain enthalpy of O has a positive value. How do you explain the existence of Mg2+02- rather than Mg+O-?
Answer:

The lattice energy of an ionic crystal depends on the force of attraction between the cations and anions \(\left(F \propto \frac{q_1 q_2}{r^2}\right)\)

So, the magnitude of lattice energy increases as the charges on the cation and anion increase. Consequently, the lattice energy of Mg²+O² is very much greater than that of Mg+O¯.

The lattice energy of Mg²+O²¯ is so high that it exceeds the unfavorable effects of the second ionization enthalpy of Mg and the second electron-gain enthalpy of O.

So, Mg²+O²¯ is a stable ionic compound and its formation is favored over Mg+O¯.

class 11 chemistry periodic classification solutions

Question 10. The atomic numbers of some elements are given below. Classify them into three groups so that the two elements in each group exhibit identical chemical behavior: 9, 12, 16, 34, 53, 56.
Answer:

Elements with atomic numbers 9 and 53 belong to the -block and they have similar outer electronic configurations (ns2np5). So they will exhibit similar chemical properties. Their group number = (10 + 2 + 5) = 17. Elements with atomic numbers 12 and 56 belong to s -block and theyhave similar outer electronic configurations (ns2).

class 11 chemistry chapter 3 exercise

So they will exhibit similar chemical properties. Their group number = 2. Elements with atomic numbers 16 and 34 belong to the -block and they have similar outer electronic configurations (ns2npA).

So, they will exhibit similar chemical properties. Their group number = (10 + 2 + 4) = 16. So, based on the similarity in chemical properties the given elements are divided into three groups :