CBSE Class 11 Chemistry Equilibrium Worksheet Set 06

Chemical Equilibrium Worksheet

Question 1. What is meant by the dynamic nature of chemical equilibrium? Describe an experiment that supports it.
Answer:
At equilibrium the forward and backward reactions have not stopped. They continue at equal rates, so the concentrations stay constant while molecules keep changing from reactants to products and back.
Experiment. Let N₂ + 3H₂ ⇌ 2NH₃ reach equilibrium. Separately, let N₂ + 3D₂ ⇌ 2ND₃ reach equilibrium under the same conditions. When the two equilibrium mixtures are mixed, NH₃, NH₂D, NHD₂ and ND₃ (and H₂, HD and D₂) all appear after some time. This shows that the reactions are still going on at equilibrium.
In simple words: Nothing looks different at equilibrium, but the molecules keep reacting in both directions at the same speed.

Teacher's Note:
1. Say clearly that the rates are equal and not zero.
2. The deuterium experiment is the standard proof in NCERT.

Question 2. State four characteristics of chemical equilibrium.
Answer:
1. The concentrations of all reactants and products remain constant at a given temperature, and the rate of the forward reaction equals the rate of the backward reaction.
2. The equilibrium is dynamic, not static.
3. It can be attained from either direction and needs a closed system. If a gas escapes or a solid separates, true equilibrium is not reached.
4. A catalyst does not change the equilibrium composition or K. It only helps to reach equilibrium sooner.
5. The value of K depends only on temperature.
In simple words: Constant amounts, equal rates, a closed vessel and no effect from a catalyst.

Teacher's Note:
1. Learn the characteristics as a short list.
2. Mention 'closed system' to earn the mark that students usually miss.

Question 3. Write the expressions for K_c (and K_p where it is meaningful) for: (a) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g); (b) CaCO₃(s) ⇌ CaO(s) + CO₂(g); (c) Zn(s) + Cu²⁺(aq) ⇌ Zn²⁺(aq) + Cu(s).
Answer:
(a) \( K_c = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5} \) and \( K_p = \frac{p_{NO}^4\,p_{H_2O}^6}{p_{NH_3}^4\,p_{O_2}^5} \)
(b) Pure solids are left out. K_c = [CO₂] and K_p = p(CO₂).
(c) K_c = [Zn²⁺] ÷ [Cu²⁺]. K_p is not used because no gas takes part.
In simple words: Products over reactants, each raised to its coefficient, and leave out pure solids and liquids.

Teacher's Note:
1. Check the coefficients before writing the powers.
2. A pure solid has a constant concentration, which is built into K.

Question 4. Derive the relation K_p = K_c(RT)^Δn for aA(g) + bB(g) ⇌ cC(g) + dD(g). What is the value of Δn for (i) N₂ + 3H₂ ⇌ 2NH₃ and (ii) PCl₅ ⇌ PCl₃ + Cl₂?
Answer:
For an ideal gas p = (n ÷ V)RT = [gas]RT, so p_A = [A]RT, and similarly for B, C and D.
K_p = (p_C)^c(p_D)^d ÷ (p_A)^a(p_B)^b = {[C]^c[D]^d ÷ [A]^a[B]^b} × (RT)^{(c + d) − (a + b)}
So K_p = K_c(RT)^Δn, where Δn = (c + d) − (a + b) = moles of gaseous products − moles of gaseous reactants.
(i) Δn = 2 − 4 = −2. (ii) Δn = 2 − 1 = +1.
In simple words: Pressure is concentration times RT, so each gas contributes a factor of RT.

Teacher's Note:
1. Count only gaseous species in Δn.
2. When Δn = 0, K_p = K_c.

Question 5. For N₂O₄(g) ⇌ 2NO₂(g), K_c = 4.6 × 10⁻³ at 298 K. Calculate K_p. (R = 0.0831 L bar K⁻¹ mol⁻¹)
Answer:
Δn = 2 − 1 = +1.
K_p = K_c(RT)^Δn = 4.6 × 10⁻³ × 0.0831 × 298
K_p = 4.6 × 10⁻³ × 24.76 = 0.114 bar
In simple words: Multiply K_c by RT once because Δn is +1.

Teacher's Note:
1. Use R = 0.0831 when pressure is in bar.
2. Change the temperature to kelvin first.

Question 6. For CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), K_c = 9 at a certain temperature. 1.0 mol L⁻¹ each of CO and H₂O is taken initially. Find the equilibrium concentrations of all four species.
Answer:
Let x mol L⁻¹ of CO react. At equilibrium [CO] = [H₂O] = 1 − x and [CO₂] = [H₂] = x.
K_c = x² ÷ (1 − x)² = 9, so x ÷ (1 − x) = 3 (take the positive root).
x = 3 − 3x, so 4x = 3 and x = 0.75.
[CO₂] = [H₂] = 0.75 mol L⁻¹ and [CO] = [H₂O] = 0.25 mol L⁻¹.
Check: (0.75 × 0.75) ÷ (0.25 × 0.25) = 9. ✓
In simple words: When both sides are perfect squares, take the square root before solving for x.

Teacher's Note:
1. Always reject a root that makes a concentration negative.
2. Substitute back to check K_c.

Question 7. For H₂(g) + I₂(g) ⇌ 2HI(g), K_c = 50 at 721 K. A mixture contains 0.5 mol L⁻¹ H₂, 0.2 mol L⁻¹ I₂ and 2.0 mol L⁻¹ HI. Is the mixture at equilibrium? If not, in which direction will the reaction proceed?
Answer:
Q_c = [HI]² ÷ ([H₂][I₂]) = (2.0)² ÷ (0.5 × 0.2) = 4.0 ÷ 0.1 = 40.
Q_c ≠ K_c, so the mixture is not at equilibrium.
Q_c (40) < K_c (50), so the reaction proceeds in the forward direction and more HI forms.
In simple words: If the quotient is smaller than K, there are too many reactants, so the reaction goes forward.

Teacher's Note:
1. Q < K: forward. Q > K: backward. Q = K: equilibrium.
2. Q has the same form as K but uses the present concentrations.

Question 8. 0.20 mol of PCl₅ is placed in a 2.0 L flask at a temperature where K_c = 0.05 mol L⁻¹ for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Find the equilibrium concentrations and the degree of dissociation.
Answer:
Initially [PCl₅] = 0.20 ÷ 2.0 = 0.10 mol L⁻¹. Let x mol L⁻¹ dissociate: [PCl₅] = 0.10 − x and [PCl₃] = [Cl₂] = x.
K_c = x² ÷ (0.10 − x) = 0.05, so x² + 0.05x − 0.005 = 0 and x = 0.05 (the positive root).
[PCl₃] = [Cl₂] = 0.05 mol L⁻¹ and [PCl₅] = 0.05 mol L⁻¹.
Degree of dissociation α = 0.05 ÷ 0.10 = 0.5 (50%).
In simple words: Half of the PCl₅ breaks up, so the three species end up with equal concentrations.

Teacher's Note:
1. Use concentrations, not moles, because Δn is not zero.
2. α = fraction of the initial amount that has dissociated.

Question 9. The equilibrium constant for N₂(g) + O₂(g) ⇌ 2NO(g) is 16 at a given temperature. Find K for (a) 2NO(g) ⇌ N₂(g) + O₂(g) and (b) ½N₂(g) + ½O₂(g) ⇌ NO(g).
Answer:
(a) The equation is reversed, so K′ = 1 ÷ K = 1 ÷ 16 = 0.0625.
(b) The equation is multiplied by ½, so K″ = K^½ = √16 = 4.
In simple words: Reverse: take the reciprocal. Halve the equation: take the square root.

Teacher's Note:
1. For multiplication by n the new constant is Kⁿ.
2. When two equations are added, multiply their K values.

Question 10. In a closed vessel CaCO₃(s) ⇌ CaO(s) + CO₂(g) is at equilibrium. What happens to the equilibrium pressure of CO₂ if (i) more CaCO₃ is added, (ii) some CO₂ is pumped out, and (iii) the temperature is raised? (The reaction is endothermic.)
Answer:
K_p = p(CO₂), so at a given temperature the equilibrium pressure of CO₂ is fixed.
(i) Adding CaCO₃ (a pure solid) has no effect.
(ii) If CO₂ is removed, the equilibrium shifts forward and more CaCO₃ decomposes until p(CO₂) returns to the same value.
(iii) For an endothermic reaction a higher temperature increases K_p, so the equilibrium pressure of CO₂ increases.
In simple words: A solid does not appear in K, so only the temperature can change the CO₂ pressure at equilibrium.

Teacher's Note:
1. In an open vessel CO₂ escapes and the decomposition goes to completion (lime kilns).
2. Only temperature changes K.

Question 11. For the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92.38 kJ mol⁻¹, state the effect on the equilibrium yield of NH₃ of (i) increasing the pressure, (ii) raising the temperature, (iii) adding a catalyst, (iv) removing NH₃ as it forms, (v) adding argon at constant volume.
Answer:
(i) Increases the yield, because the number of gas moles falls from 4 to 2.
(ii) Decreases the yield, because the forward reaction is exothermic (K falls). A moderate temperature of about 700 K is used to get a reasonable rate.
(iii) No change in the yield. The catalyst only speeds up the approach to equilibrium.
(iv) Increases the yield, as the equilibrium shifts forward.
(v) No change, because the concentrations of the reacting gases stay the same.
In simple words: High pressure, removal of NH₃ and low temperature raise the yield; a catalyst only speeds things up.

Teacher's Note:
1. Explain each answer with Le Chatelier's principle.
2. Industry compromises between yield and rate.

Question 12. Calculate ΔG° at 298 K for a reaction with K = 10. (R = 8.314 J K⁻¹ mol⁻¹, ln 10 = 2.303.) What does the sign tell you?
Answer:
ΔG° = −RT ln K = −2.303 RT log K
ΔG° = −8.314 × 298 × 2.303 = −5706 J mol⁻¹ = −5.71 kJ mol⁻¹.
ΔG° is negative because K > 1, so products are favoured at equilibrium and the forward reaction is spontaneous under standard conditions.
In simple words: A negative standard Gibbs energy change goes with a K larger than 1.

Teacher's Note:
1. Convert J to kJ.
2. If K < 1, ΔG° is positive.

Question 13. The reaction 2NO₂(g) ⇌ N₂O₄(g), ΔH = −57.2 kJ mol⁻¹, is carried out in a sealed tube. How does the brown colour of the gas change on (i) heating and (ii) cooling? Explain.
Answer:
NO₂ is brown and N₂O₄ is colourless. The forward reaction is exothermic.
(i) On heating, the equilibrium shifts backward (to absorb heat) and more NO₂ forms. The brown colour deepens.
(ii) On cooling, the equilibrium shifts forward, more N₂O₄ forms and the colour fades.
In simple words: Heat favours the endothermic backward reaction, so more brown NO₂ appears.

Teacher's Note:
1. For an exothermic reaction, K decreases when the temperature rises.
2. This is a good demonstration of Le Chatelier's principle.

Question 14. Predict the extent of reaction from K_c in each case: (a) K_c = 5 × 10⁻³⁹, (b) K_c = 3.7 × 10⁸, (c) K_c = 1.8.
Answer:
(a) K_c < 10⁻³: the reaction hardly proceeds, and the equilibrium mixture is mostly reactants.
(b) K_c > 10³: the reaction goes almost to completion, and products dominate.
(c) K_c between 10⁻³ and 10³: appreciable amounts of both reactants and products are present.
K_c says nothing about the rate of reaching equilibrium.
In simple words: A very large K means mostly products, a very small K means mostly reactants.

Teacher's Note:
1. Compare K with 10³ and 10⁻³.
2. A large K does not mean a fast reaction.