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Vidaara.orgClass 11 · Chemistry
CodeVID-C11-07-CH-01
Chapter Assignment — Equilibrium
Chapter: Equilibrium
Topic: All Topics
Maximum Marks: 40
Time: 90 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • This is a full-length test covering the whole chapter — every topic is included.
  • 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.
Section A — Multiple Choice Questions 6 × 1 = 6 marks
1.
At equilibrium the rates of the forward and reverse reactions are:
  • A.zero
  • B.equal
  • C.forward greater
  • D.reverse greater
2.
For $\Delta n=0$, the relation $K_p$ 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.
The conjugate acid of $OH^-$ is:
  • A.$O^{2-}$
  • B.$H_2O$
  • C.$H_3O^+$
  • D.$H_2O_2$
4.
The pH of $0.01\ \text{M}$ NaOH is:
  • A.2
  • B.7
  • C.12
  • D.14
5.
For a salt $AB$ of solubility $s$, $K_{sp}=$
  • A.$s$
  • B.$s^2$
  • C.$4s^3$
  • D.$27s^4$
6.
A precipitate forms when the ionic product is:
  • A.less than $K_{sp}$
  • B.equal to $K_{sp}$
  • C.greater than $K_{sp}$
  • D.zero
Section B — Short Answer (2 marks) 4 × 2 = 8 marks
7.
State Le Chatelier's principle.
8.
Write the conjugate base of $HCO_3^-$ and its conjugate acid.
9.
Calculate the pH of $0.005\ \text{M}$ $H_2SO_4$ (complete ionisation).
10.
Define solubility product and write it for $CaF_2$ in terms of $s$.
Section C — Short Answer (3 marks) 2 × 3 = 6 marks
11.
For $PCl_5\rightleftharpoons PCl_3+Cl_2$, $1\ \text{mol}$ in a $1\ \text{L}$ flask gives $0.6\ \text{mol}$ of $Cl_2$ at equilibrium. Find $K_c$.
12.
Calculate the pH of a buffer of $0.1\ \text{M}$ acetic acid and $0.1\ \text{M}$ sodium acetate ($pK_a=4.74$), then state the effect of diluting it.
Section D — Long Answer (5 marks) 2 × 5 = 10 marks
13.
(a) Derive $K_p=K_c(RT)^{\Delta n}$. (b) For $2SO_2+O_2\rightleftharpoons 2SO_3$ ($\Delta H<0$), use Le Chatelier's principle to discuss the effect of pressure, temperature and a catalyst on the yield of $SO_3$.
14.
Define $K_{sp}$ and the common ion effect. The $K_{sp}$ of $AgCl$ is $1.8\times10^{-10}$. (a) Find its solubility in pure water. (b) Find its solubility in $0.1\ \text{M}$ $NaCl$ and comment.

Answer Key

Section A — Multiple Choice Questions
  1. (B) equal
  2. (B) $K_p=K_c$
  3. (B) $H_2O$
  4. (C) 12
  5. (B) $s^2$
  6. (C) greater than $K_{sp}$
Section B — Short Answer (2 marks)
  1. If a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that opposes (counteracts) the disturbance.
  2. Conjugate base (lose $H^+$): $CO_3^{2-}$. Conjugate acid (gain $H^+$): $H_2CO_3$. Thus $HCO_3^-$ is amphoteric.
  3. Each mole gives 2 $H^+$, so $[H^+]=0.01\ \text{M}$; $pH=-\log(0.01)=2$.
  4. $K_{sp}$ is the product of molar ion concentrations (each raised to its coefficient) in a saturated solution at equilibrium. For $CaF_2\rightleftharpoons Ca^{2+}+2F^-$: $K_{sp}=s(2s)^2=4s^3$.
Section C — Short Answer (3 marks)
  1. At eq: $PCl_3=Cl_2=0.6$, $PCl_5=0.4\ \text{mol L}^{-1}$; $K_c=\dfrac{0.6\times0.6}{0.4}=\dfrac{0.36}{0.4}=0.9\ \text{mol L}^{-1}$.
  2. $pH=pK_a+\log\dfrac{[salt]}{[acid]}=4.74+\log 1=4.74$. On dilution the ratio $[salt]/[acid]$ is unchanged, so the pH stays essentially $4.74$.
Section D — Long Answer (5 marks)
  1. (a) Using $p_i=[i]RT$ for ideal gases, substitute into $K_p$: $K_p=\dfrac{[SO_3]^2(RT)^2}{[SO_2]^2(RT)^2[O_2](RT)}=K_c(RT)^{2-3}=K_c(RT)^{\Delta n}$ with $\Delta n=-1$. In general $\Delta n$ = (moles of gaseous products) - (moles of gaseous reactants). (b) Pressure: 3 mol reactant gas vs 2 mol product gas, so high pressure shifts forward, increasing $SO_3$. Temperature: the forward reaction is exothermic, so low temperature increases $K$ and yield (but rate falls, so a moderate ~720 K is used industrially). Catalyst ($V_2O_5$): speeds attainment of equilibrium but does not change the yield.
  2. $K_{sp}$ is the equilibrium ion product of a saturated solution of a sparingly soluble salt; the common ion effect is the suppression of ionisation/solubility on adding an ion already present. (a) In pure water $K_{sp}=s^2$, so $s=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}\ \text{mol L}^{-1}$. (b) In $0.1\ \text{M}$ $NaCl$, $[Cl^-]\approx0.1$, so $s'=\dfrac{K_{sp}}{[Cl^-]}=\dfrac{1.8\times10^{-10}}{0.1}=1.8\times10^{-9}\ \text{mol L}^{-1}$. The solubility falls by about $10^4$ times — a clear demonstration of the common ion effect.
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