Vidaara.orgClass 11 · Chemistry
CodeVID-C11-07-T1-01
Assignment — Physical & Chemical Equilibrium
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Equilibrium in a chemical reaction is best described as:
- A.static
- B.dynamic
- C.irreversible
- D.instantaneous
2.
For a reaction with $\Delta n=0$, the relation between $K_p$ and $K_c$ is:
- A.$K_p=K_c RT$
- B.$K_p=K_c$
- C.$K_p=K_c/RT$
- D.$K_p=K_c(RT)^2$
3.
A very large equilibrium constant indicates:
- A.reactants dominate
- B.products dominate
- C.reaction does not occur
- D.equal amounts of both
4.
Increasing pressure on $2SO_2+O_2\rightleftharpoons 2SO_3$ shifts equilibrium:
- A.forward
- B.reverse
- C.no change
- D.stops the reaction
5.
Only this factor can change the numerical value of $K$:
- A.concentration
- B.pressure
- C.temperature
- D.catalyst
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State the law of mass action and write $K_c$ for $aA+bB\rightleftharpoons cC+dD$.
7.
Explain why a catalyst does not change the equilibrium position.
8.
Define the reaction quotient $Q$ and state how it predicts the direction of reaction.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
For $H_2+I_2\rightleftharpoons 2HI$ a $1\ \text{L}$ flask at equilibrium has $0.5\ \text{mol}$ $H_2$, $0.5\ \text{mol}$ $I_2$ and $2\ \text{mol}$ $HI$. Find $K_c$.
10.
State Le Chatelier's principle and apply it to the effect of (i) removing a product and (ii) raising temperature on an endothermic reaction.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Derive the relation $K_p=K_c(RT)^{\Delta n}$ for a gaseous equilibrium and explain the meaning of $\Delta n$.
Answer Key
Section A — Multiple Choice Questions
- (B) dynamic
- (B) $K_p=K_c$
- (B) products dominate
- (A) forward
- (C) temperature
Section B — Short Answer (2 marks)
- Rate is proportional to the product of active masses raised to stoichiometric coefficients; $K_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$.
- A catalyst lowers the activation energy of the forward and reverse reactions equally, so it speeds both rates by the same factor; equilibrium is reached sooner but its position and $K$ are unchanged.
- $Q$ has the same form as $K$ but uses current concentrations. If $Q
Section C — Short Answer (3 marks)
- $K_c=\dfrac{[HI]^2}{[H_2][I_2]}=\dfrac{(2)^2}{0.5\times 0.5}=\dfrac{4}{0.25}=16$.
- A system at equilibrium shifts to oppose any disturbance. (i) Removing a product shifts the equilibrium forward to replace it. (ii) Raising $T$ favours the endothermic (forward) direction and increases $K$.
Section D — Long Answer (5 marks)
- For a gas $p_i=\dfrac{n_i}{V}RT=[i]RT$ (ideal gas). Substituting partial pressures into $K_p$ and concentrations into $K_c$ for $aA+bB\rightleftharpoons cC+dD$: $K_p=\dfrac{(p_C)^c(p_D)^d}{(p_A)^a(p_B)^b}=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}(RT)^{(c+d)-(a+b)}=K_c(RT)^{\Delta n}$, where $\Delta n=(c+d)-(a+b)$ is the change in number of gaseous moles. When $\Delta n=0$, $K_p=K_c$.
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