JEE Chemistry Chemical Kinetics MCQs Set 17

Answer: (a) \( -\dfrac{1}{3}\dfrac{d[C]}{dt} \)

Explanation:
1. Rate = \( -\dfrac{1}{2}\dfrac{d[A]}{dt} = -\dfrac{d[B]}{dt} = \dfrac{1}{3}\dfrac{d[C]}{dt} = \dfrac{d[D]}{dt} \).
2. Options b, c and d match this.
3. The product C carries a positive sign, so \( -\dfrac{1}{3}\dfrac{d[C]}{dt} \) does not express the rate.

Teacher's Note:
1. Divide each rate of change by its coefficient.
2. Reactants take a minus sign and products a plus sign.

Answer: (c) \( \dfrac{1000}{2.303} \) K

Explanation:
1. Put \( k_1 = k_2 \): \( 10^{16}e^{-2000/T} = 10^{15}e^{-1000/T} \), so \( 10 = e^{1000/T} \).
2. Taking natural logarithms, \( 1000/T = \ln 10 = 2.303 \).
3. Hence \( T = 1000/2.303 \) K.

Teacher's Note:
1. Divide the two expressions and take the natural logarithm.
2. ln 10 = 2.303.

Answer: (d) Rate = k[CH3COCH3][H+]

Explanation:
1. Rows 1 and 2: doubling [Br2] does not change the rate, so the order in Br2 is zero.
2. Rows 2 and 3: doubling [H+] roughly doubles the rate (5.7 × 10⁻⁵ to 1.2 × 10⁻⁴), so the order in H+ is one.
3. Rows 1 and 4: acetone increases by 4/3 and [H+] by 4, so the rate should rise about 5.3 times, and 3.1 × 10⁻⁴ / 5.7 × 10⁻⁵ = 5.4 agrees. So the order in acetone is one.

Teacher's Note:
1. Compare two rows in which only one concentration changes.
2. A concentration that does not affect the rate has order zero.

Answer: (c) rate = k[A][B]²

Explanation:
1. Doubling A alone doubles the rate, so the order in A is 1.
2. Doubling both gives \( 2^{1} \times 2^{n} = 8 \), so \( 2^n = 4 \) and the order in B is 2.
3. Rate = k[A][B]².

Teacher's Note:
1. Find the order in one reactant first and then use the combined change.
2. The total factor is the product of the individual factors.

Answer: (b) \( \dfrac{d[Br_2]}{dt} = -\dfrac{3}{5}\dfrac{d[Br^-]}{dt} \)

Explanation:
1. Rate = \( \dfrac{1}{3}\dfrac{d[Br_2]}{dt} = -\dfrac{1}{5}\dfrac{d[Br^-]}{dt} \).
2. So \( \dfrac{d[Br_2]}{dt} = -\dfrac{3}{5}\dfrac{d[Br^-]}{dt} \).

Teacher's Note:
1. Bromine forms 3 units for every 5 units of bromide used.
2. The minus sign shows that bromide is being consumed.

Answer: (d) \( 0.5 \times 10^{-3} \) s⁻¹

Explanation:
1. \( k = 0.693/t_{1/2} = 0.693/1386 \).
2. \( k = 5 \times 10^{-4} \) s⁻¹ = \( 0.5 \times 10^{-3} \) s⁻¹.

Teacher's Note:
1. For first order, \( k = 0.693/t_{1/2} \).
2. 1386 = 2 × 693, which makes the arithmetic easy.

Answer: (a) rate = k[A][B]²

Explanation:
1. Experiments II and III: [A] constant, [B] doubles, rate becomes 4 times, so the order in B is 2.
2. Experiments I and IV: [B] constant, [A] becomes 4 times and rate becomes 4 times, so the order in A is 1.
3. Rate = k[A][B]². Check with I and II: 3 × 2² = 12 times, and 6.0 × 10⁻³ × 12 = 7.2 × 10⁻².

Teacher's Note:
1. Always pick two experiments in which only one concentration changes.
2. Verify the result with a third pair.

Answer: (b) \( \dfrac{1}{t}\ln\dfrac{P_0}{2P_0 - P_t} \)

Explanation:
1. If A falls by x, B rises by 2x. Solid C does not contribute to pressure.
2. \( P_t = (P_0 - x) + 2x = P_0 + x \), so \( x = P_t - P_0 \).
3. \( P_A = P_0 - x = 2P_0 - P_t \), so \( k = \dfrac{1}{t}\ln\dfrac{P_0}{2P_0 - P_t} \).

Teacher's Note:
1. Count only the gaseous species when you write the total pressure.
2. Pressure of A is the initial pressure minus the part that reacted.

Answer: (d) Order of a reaction is always a whole number

Explanation:
1. Order comes only from experiment and is the sum of powers in the rate law, whatever the stoichiometric coefficients.
2. It can be zero, fractional or even negative, so it is not always a whole number.

Teacher's Note:
1. Molecularity is always a whole number but order need not be.
2. Order and stoichiometry are generally unrelated.

Answer: (d) 690 s

Explanation:
1. The ratio 1.28/0.04 = 32 = 2⁵, so five half-lives are needed.
2. Time = 5 × 138 = 690 s.

Teacher's Note:
1. Express the fall in concentration as a power of 2.
2. The exponent is the number of half-lives.

Answer: (c) k' = 2k; k'' = k/2

Explanation:
1. From the stoichiometry \( \dfrac{d[NO_2]}{dt} = 2\left(-\dfrac{d[N_2O_5]}{dt}\right) \), so k' = 2k.
2. \( \dfrac{d[O_2]}{dt} = \dfrac{1}{2}\left(-\dfrac{d[N_2O_5]}{dt}\right) \), so k'' = k/2.

Teacher's Note:
1. Two moles of N2O5 give four moles NO2 and one mole O2.
2. So NO2 forms at twice the N2O5 rate and O2 at half of it.

Answer: (b) \( \dfrac{2.303}{t}\log\dfrac{2P_0}{3P_0 - P_t} \)

Explanation:
1. If A falls by x, then B = 2x and C = x.
2. \( P_t = P_0 - x + 2x + x = P_0 + 2x \), so \( x = (P_t - P_0)/2 \).
3. \( P_A = P_0 - x = (3P_0 - P_t)/2 \), so \( k = \dfrac{2.303}{t}\log\dfrac{2P_0}{3P_0 - P_t} \).

Teacher's Note:
1. Total pressure is the sum of the partial pressures of all gases.
2. Here three moles of gas are produced from one mole of A.

Answer: (a) mol L⁻¹ s⁻¹ and mol^{-1/2} L^{1/2} s⁻¹

Explanation:
1. Rate always has the unit mol L⁻¹ s⁻¹.
2. \( k = \text{rate}/c^{3/2} \), so the unit of k is \( (\text{mol L}^{-1}\text{ s}^{-1})/(\text{mol L}^{-1})^{3/2} \).
3. This gives mol^{-1/2} L^{1/2} s⁻¹.

Teacher's Note:
1. Unit of k is (mol L⁻¹)^(1 − n) s⁻¹.
2. Here n = 3/2, so the power is −1/2.

Answer: (d) third order

Explanation:
1. \( t_{1/2} \propto a^{1-n} \).
2. A straight line against \( 1/a^2 = a^{-2} \) means \( 1 - n = -2 \).
3. So n = 3.

Teacher's Note:
1. Use \( t_{1/2} \propto a^{1-n} \).
2. Match the power of a with the plotted quantity.

Answer: (d) \( \Delta H = 0 \)

Explanation:
1. \( \Delta H = E_a(\text{forward}) - E_a(\text{reverse}) \).
2. Equal activation energies give \( \Delta H = 0 \).
3. Nothing can be said about \( \Delta G \) or \( \Delta S \) from this.

Teacher's Note:
1. The enthalpy change equals the difference of the two activation energies.
2. Entropy and free energy are not fixed by this condition.

Answer: (b) 6 h

Explanation:
1. 75% completion is two half-lives, so \( t_{1/2} = 2 \) h.
2. 87.5% completion is three half-lives.
3. Time = 3 × 2 = 6 h.

Teacher's Note:
1. Fraction left: 25% = (1/2)², 12.5% = (1/2)³.
2. Count half-lives to get the time.

Answer: (a) first

Explanation:
1. \( t_{1/2} \propto a^{1-n} \).
2. A half-life independent of a requires \( 1 - n = 0 \), so n = 1.

Teacher's Note:
1. Only first order reactions have a constant half-life.
2. Zero order half-life rises with a and second order half-life falls.

Answer: (b) 0.72 M

Explanation:
1. The unit of k shows a zero order reaction.
2. Amount reacted = kt = 0.6 × 10⁻³ × (20 × 60) = 0.72 M.
3. The concentration of B formed is 0.72 M (A is far from used up).

Teacher's Note:
1. The unit mol L⁻¹ s⁻¹ identifies zero order.
2. Convert minutes to seconds before multiplying.

Answer: (b) 1000 K

Explanation:
1. \( k = 0.693/t_{1/2} = 0.693/(6.93 \times 10^{-3}) = 100 \) min⁻¹.
2. \( \log k = 2 \), so \( 14 - 1.2 \times 10^4/T = 2 \).
3. \( T = 1.2 \times 10^4/12 = 1000 \) K.

Teacher's Note:
1. Get k from the half-life first.
2. Then substitute into the given log equation.

Answer: (a) 0

Explanation:
1. \( t_{1/2} \propto a^{1-n} \).
2. Halving a halves \( t_{1/2} \), so \( (1/2)^{1-n} = 1/2 \).
3. Hence 1 − n = 1 and n = 0.

Teacher's Note:
1. Half-life that falls in proportion to concentration means zero order.
2. Use \( t_{1/2} \propto a^{1-n} \).

Answer: (a) 24.1 s

Explanation:
1. For first order, rate is proportional to concentration, so \( k = \dfrac{2.303}{10}\log\dfrac{0.04}{0.03} \).
2. \( k = 0.2303 \times 0.1249 = 0.02877 \) s⁻¹.
3. \( t_{1/2} = 0.693/0.02877 = 24.1 \) s.

Teacher's Note:
1. The ratio of rates equals the ratio of concentrations in a first order reaction.
2. The gap in time is 10 s.

Answer: (d) Activation energy

Explanation:
1. A catalyst gives the reaction a new path of lower activation energy.
2. State functions such as entropy, internal energy and enthalpy depend only on the initial and final states, so they stay the same.

Teacher's Note:
1. A catalyst changes the path and not the initial or final state.
2. Only the activation energy is altered.

Answer: (b) \( 2E_A = E_B \)

Explanation:
1. \( \ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right) \) with the same temperature interval, so \( E_a \propto \ln(k_2/k_1) \).
2. \( \ln 9 = 2\ln 3 \), so \( E_B = 2E_A \).

Teacher's Note:
1. For the same temperature change, Ea is proportional to ln of the rate ratio.
2. 9 is 3 squared.