A_{(g)} xrightarrow{Delta} P_{(g)} + Q_{(g)} | vedclass

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(B) Given,half-life $t_{1/2} = 69.3 \ s$.For a first-order reaction,the rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{69.3} = 10^{-2} \ s^{-

Vedclass · Accepted Answer

$1.12$

Vedclass · Answer

(B) Given,half-life $t_{1/2} = 69.3 \ s$.For a first-order reaction,the rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{69.3} = 10^{-2} \ s^{-1}$.For first-order kinetics,$k = \frac{2.303}{t} \log \frac{P_0}{P_t}$,where $P_0$ is initial pressure and $P_t$ is pressure at time $t$.$10^{-2} = \frac{2.303}{230} \log \frac{0.4}{P_t}$.$1 = \log \frac{0.4}{P_t} \implies \frac{0.4}{P_t} = 10 \implies P_t = 0.04 \ atm$.From the reaction $A_{(g)} \rightarrow P_{(g)} + Q_{(g)} + R_{(g)}$:At $t = 0$,$P_A = 0.4 \ atm$,$P_P = 0, P_Q = 0, P_R = 0$.At $t = 230 \ s$,$P_A = 0.04 \ atm$. The decrease in pressure of $A$ is $0.4 - 0.04 = 0.36 \ atm$.According to stoichiometry,the pressure of products formed is $P_P = 0.36 \ atm, P_Q = 0.36 \ atm, P_R = 0.36 \ atm$.Total pressure $P_{total} = P_A + P_P + P_Q + P_R = 0.04 + 0.36 + 0.36 + 0.36 = 1.12 \ atm$.

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