JEE Chemistry Chemical Equilibrium MCQs Set 04

Answer: (c) \( \frac{b+a\alpha}{a+b+c+a\alpha}\,P \) atm

Explanation:
1. At equilibrium: PCl5 = \( a-a\alpha \), PCl3 = \( b+a\alpha \), Cl2 = \( c+a\alpha \).
2. Total moles = \( a+b+c+a\alpha \).
3. Partial pressure = mole fraction x total pressure, so \( p_{PCl_3}=\frac{b+a\alpha}{a+b+c+a\alpha}P \).

Teacher's Note:
1. Always count the moles of every species at equilibrium, including those present initially, before forming mole fractions.
2. The expression \( a(1-\alpha) \) in the numerator would be the partial pressure of PCl5, not PCl3.

Answer: (a) 0.512 atm\(^3\)

Explanation:
1. S8 falls from 1 atm by 0.2 atm, so \( p_{S_8}=0.8 \) atm.
2. For every S8 lost, four S2 form: \( p_{S_2}=4\times0.2=0.8 \) atm.
3. \( K_p=\frac{(0.8)^4}{0.8}=0.8^3=0.512 \) atm\(^3\).

Teacher's Note:
1. The units of \( K_p \) follow from the change in moles of gas: here \( \Delta n=3 \), so the unit is atm\(^3\).
2. Use the stoichiometric ratio to convert the loss of reactant to the gain of product.

Answer: (d) 0.2704

Explanation:
1. Only the gases appear in \( K_p \); CO2 and H2O are formed in equal amounts, so each has a partial pressure of \( 1.04/2=0.52 \) atm.
2. \( K_p=p_{CO_2}\times p_{H_2O}=0.52\times0.52=0.2704 \).

Teacher's Note:
1. Pure solids are left out of the equilibrium constant expression.
2. When two gases come from one solid in a 1 : 1 ratio, each contributes half of the total pressure.

Answer: (b) \( \frac{1}{1+\sqrt{2}} \)

Explanation:
1. Start with 1 mol each of A2 and B2 in volume V and let x mol of each react: A2 = B2 = \( 1-x \), AB = \( 2x \).
2. \( K=\frac{(2x)^2}{(1-x)^2}=2 \), so \( \frac{2x}{1-x}=\sqrt{2} \).
3. \( x=\frac{\sqrt{2}}{2+\sqrt{2}}=\frac{1}{\sqrt{2}+1} \approx 0.414 \).

Teacher's Note:
1. Because the number of moles does not change, the volume cancels from K.
2. Taking the square root of both sides turns the quadratic into a simple linear equation.

Answer: (b) 66%

Explanation:
1. Let 1 mol each of acid and alcohol give x mol of ester and x mol of water.
2. \( K=\frac{x\cdot x}{(1-x)(1-x)}=4 \), so \( \frac{x}{1-x}=2 \).
3. \( x=\frac{2}{3}\approx0.667 \), i.e. about 66-67% of the alcohol is esterified.

Teacher's Note:
1. Equal moles of reactants and no change in the number of moles mean the volume cancels out.
2. When K is a perfect square, take the square root first to avoid solving a quadratic.

Answer: (d) \( \ce{N2 + O2 <=> 2NO} \)

Explanation:
1. Vapour density is inversely proportional to the volume, and so to the total number of moles of gas.
2. The other three reactions increase the number of moles on dissociation, so vapour density changes with the degree of dissociation.
3. In \( \ce{N2 + O2 <=> 2NO} \), \( \Delta n=0 \): the number of moles and the molar mass stay the same, so the vapour density does not change.

Teacher's Note:
1. The method works only when \( \Delta n_g \neq 0 \).
2. The relation used is \( \alpha=\frac{D-d}{(n-1)d} \), where D is the initial and d the observed vapour density.

Answer: (a) 9 : 1 : 1

Explanation:
1. Dissociated N2O3 = \( 0.10\times2=0.2 \) mol L\(^{-1}\).
2. [N2O3] = \( 2-0.2=1.8 \); [NO] = 0.2; [NO2] = 0.2.
3. Ratio = 1.8 : 0.2 : 0.2 = 9 : 1 : 1.

Teacher's Note:
1. Each mole of N2O3 that dissociates gives one mole of NO and one of NO2.
2. Always scale the ratio down to the smallest whole numbers.

Answer: (c) 0.573 M

Explanation:
1. Let x mol L\(^{-1}\) of CO form: [H2] = [CO2] = \( 1-x \), [H2O] = [CO] = x.
2. \( K=\frac{x^2}{(1-x)^2}=1.8 \), so \( \frac{x}{1-x}=1.34 \).
3. \( x=\frac{1.34}{2.34}=0.573 \) M.

Teacher's Note:
1. When the number of moles of gas does not change, K can be written directly in concentrations without needing the volume.
2. Taking the square root of a perfect-square expression avoids the quadratic formula.

Answer: (c) \( K_1^3\times K_2=K_3 \)

Explanation:
1. Multiply reaction I by 3: \( \ce{3A + 6B <=> 3C} \), constant \( K_1^3 \).
2. Add reaction II: \( \ce{3A + 6B + C + D <=> 3C + 3A} \), constant \( K_1^3K_2 \).
3. Cancelling 3A and one C gives \( \ce{6B + D <=> 2C} \), which is reaction III. So \( K_3=K_1^3K_2 \).

Teacher's Note:
1. Multiplying a reaction by n raises its K to the power n; adding reactions multiplies their K values.
2. Never multiply K by a coefficient: the coefficient becomes an exponent.

Answer: (a) 3.11

Explanation:
1. \( \ce{S + S^{2-} <=> S2^{2-}} \), \( K_a=1.7 \); \( \ce{2S + S^{2-} <=> S3^{2-}} \), \( K_b=5.3 \).
2. Required: \( \ce{S + S2^{2-} <=> S3^{2-}} \), obtained by subtracting the first from the second.
3. \( K=\frac{K_b}{K_a}=\frac{5.3}{1.7}=3.11 \).

Teacher's Note:
1. Subtracting one reaction from another divides its equilibrium constant.
2. Write the target reaction first and then decide which given reactions to add or subtract.

Answer: (d) III < I = IV < II

Explanation:
1. The extent of a reaction is measured by K. With every concentration 0.1 M, \( K=(0.1)^{\Delta n} \) with \( \Delta n \) = products minus reactants (in moles).
2. I: \( K=\frac{(0.1)^2}{(0.1)^3}=10 \). II: \( K=\frac{(0.1)^2}{(0.1)^4}=100 \). III: \( K=\frac{(0.1)^2}{0.1}=0.1 \). IV: \( K=\frac{(0.1)^9}{(0.1)^{10}}=10 \).
3. Order of K (and of extent): III (0.1) < I = IV (10) < II (100).

Teacher's Note:
1. A larger K means a greater extent of forward reaction.
2. Compare K values directly only for the same stoichiometric form of each reaction as written.

Answer: (b) 45

Explanation:
1. Start with 1 mol NOBr: NOBr = 0.6, NO = 0.4, Br2 = 0.2, total = 1.2 mol.
2. Partial pressures: NOBr = \( \frac{0.6}{1.2}\times0.30=0.15 \), NO = 0.10, Br2 = 0.05 atm.
3. For dissociation: \( K_p=\frac{(0.10)^2(0.05)}{(0.15)^2}=0.0222 \).
4. The reaction asked is the reverse, so \( K_p'=\frac{1}{0.0222}=45 \).

Teacher's Note:
1. Always check which direction of the reaction the question asks about: reversing a reaction inverts K.
2. The value 0.022 is the trap option: it is K for the dissociation, not for the formation of NOBr.

Answer: (b) 0.67

Explanation:
1. From the graph, at equilibrium A is 60 mole % and B is 40 mole %.
2. \( K=\frac{[B]}{[A]}=\frac{40}{60}=0.67 \).

Teacher's Note:
1. Since \( \Delta n=0 \), mole percentages can be used in place of concentrations.
2. A curve falling from 100 is the reactant; the one rising from 0 is the product.

Answer: (d) -71080.57 J mol\(^{-1}\)

Explanation:
1. Use \( \ln\frac{K_2}{K_1}=\frac{\Delta H}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \).
2. \( \ln\frac{9.25}{18.5}=-0.693 \) and \( \frac{1}{925}-\frac{1}{1000}=8.108\times10^{-5} \).
3. \( \Delta H=\frac{8.314\times(-0.693)}{8.108\times10^{-5}}\approx-7.108\times10^4 \) J mol\(^{-1}\).

Teacher's Note:
1. K falls as temperature rises, which tells you at once that the reaction is exothermic (negative \( \Delta H \)).
2. This eliminates the positive option before any calculation.

Answer: (a) \( \frac{(1-2x)^2}{(0.5+x)x} \)

Explanation:
1. There is no I2 initially, so the system moves backwards: \( \ce{2HI -> H2 + I2} \).
2. If x mol of I2 forms, then H2 = \( 0.5+x \), HI = \( 1-2x \).
3. \( K=\frac{[HI]^2}{[H2][I2]}=\frac{(1-2x)^2}{(0.5+x)\,x} \).

Teacher's Note:
1. When one species is absent initially, decide the direction of reaction before setting up the table.
2. Two moles of HI are consumed for each mole of I2 produced.

Answer: (c) \( T_1>T_2>T_3 \)

Explanation:
1. For an exothermic reaction, a lower temperature gives a higher equilibrium yield.
2. At any given pressure the yield is highest for T3 and lowest for T1.
3. Therefore T3 is the lowest temperature and T1 the highest: \( T_1>T_2>T_3 \).

Teacher's Note:
1. Rising temperature shifts an exothermic equilibrium backwards (Le Chatelier).
2. Compare curves at one fixed pressure, never across different pressures.

Answer: (c) At high pressure and high temperature

Explanation:
1. To favour ozone we must shift the equilibrium to the left.
2. The forward reaction is exothermic, so the backward reaction is endothermic and is favoured by high temperature.
3. The forward reaction increases the moles of gas (2 to 3), so the backward reaction is favoured by high pressure.

Teacher's Note:
1. Apply Le Chatelier's principle to each variable separately: temperature through the sign of \( \Delta H \), pressure through \( \Delta n_g \).
2. Ozone is stable only because high T and high P both push the equilibrium toward the side with fewer moles.

Answer: (a) 3.3, 2.3

Explanation:
1. Total pressure \( P=\frac{n}{V}RT=0.25\times0.0821\times273=5.6 \) atm.
2. \( p_{N_2}=0.59\times5.6=3.3 \) atm.
3. \( p_{O_2}=0.41\times5.6=2.3 \) atm.

Teacher's Note:
1. Use \( P=CRT \) when concentration (mol per litre) is given.
2. The two partial pressures must add up to the total pressure, 5.6 atm.

Answer: (d) 4 atm

Explanation:
1. Moles of O2 = \( 3.2/32=0.1 \); moles of H2 = \( 0.2/2=0.1 \); total = 0.2 mol.
2. \( P=\frac{nRT}{V}=\frac{0.2\times0.0821\times273}{1.12}\approx4.0 \) atm.

Teacher's Note:
1. At 0 \( ^\circ \)C (273 K) one mole of gas occupies 22.4 L at 1 atm, so 0.2 mol in 1.12 L gives \( 1\times\frac{0.2\times22.4}{1.12}=4 \) atm.
2. Moles of the two gases simply add, since they do not react at this temperature.