The first-order decomposition \( \ce{SO2Cl2(g) -> SO2(g) + Cl2(g)} \) is carried out at constant volume and constant temperature, starting with pure SO2Cl2 at a pressure of 0.40 atm. After 200 s, the total pressure of the gas mixture is 0.70 atm. What is the rate constant of the reaction? (log 2 = 0.301, log 3 = 0.477, log 7 = 0.845)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 6.93 × 10−3 s−1
In a gas-phase reaction at constant volume, partial pressure stands in for concentration. The total pressure rises because one molecule of SO2Cl2 gives two gas molecules. First find the pressure of SO2Cl2 left, then use the first-order equation.
\( \displaystyle \ce{SO2Cl2(g) -> SO2(g) + Cl2(g)} \)
At t = 0: 0.40 atm, 0, 0. At time t: (0.40 − x) atm, x atm, x atm.
\( \displaystyle p_t = (0.40 - x) + x + x = 0.40 + x = 0.70\ \mathrm{atm} \Rightarrow x = 0.30\ \mathrm{atm} \)
\( \displaystyle p_{\ce{SO2Cl2}} = 2p_i - p_t = 0.80 - 0.70 = 0.10\ \mathrm{atm} \)
\( \displaystyle k = \frac{2.303}{t}\log\frac{p_i}{p_{\ce{SO2Cl2}}} = \frac{2.303}{200\ \mathrm{s}}\log\frac{0.40}{0.10} \)
\( \displaystyle k = \frac{2.303 \times 0.602}{200\ \mathrm{s}} = 6.93 \times 10^{-3}\ \mathrm{s^{-1}} \)
The pressure of SO2Cl2 fell from 0.40 to 0.10 atm, to one-fourth, which is two half-lives in 200 s; so t1/2 = 100 s and k = 0.693/100 s gives the same value.
NCERT (2026-27 edition):
3.3.2 First Order Reactions – “Let pi be the initial pressure of A and pt the total pressure at time ‘t’. … If x atm be the decrease in pressure of A at time t and one mole each of B and C is being formed, the increase in pressure of B and C will also be x atm each.” … “pt = (pi – x) + x + x = pi + x … x = (pt – pi) where, pA = pi – x = pi – (pt – pi) = 2pi – pt”
\( \displaystyle k = \frac{2.303}{t}\log\frac{p_i}{p_A} = \frac{2.303}{t}\log\frac{p_i}{(2p_i - p_t)} \qquad (3.16) \)