Q64 / 180 · AIIMS 2014 · Chemistry

For the reaction $A_{(g)} \longrightarrow B_{(g)}+C_{(g)}$, the rate constant is given as $\left(P_i\right.$ is initial pressure and $P_t$ is pressure at time $t$ )

  • A
    $k=\frac{2.303}{t} \log \frac{P_i}{P_t}$
  • Correct answer
    B
    $k=\frac{2.303}{t} \log \frac{P_i}{\left(2 P_i-P_t\right)}$
  • C
    $k=\frac{2.303}{t} \log \frac{2 P_i-P_t}{P_i}$
  • D
    $k=\frac{2.303}{t} \log \frac{P_i-P_t}{2 P_i}$

Solution

For the reaction, $A_{(g)} \longrightarrow B_{(g)}+C_{(g)}$ $\begin{array}{llll} t=0 & P_i & 0 & 0 \\ t & \left(P_i-x\right) & x & x \end{array}$ Given: $P_i$ is initial pressure at time, $t=0$ and $P_t$ is pressure at time ' $t$ '. $P_t=\left(P_i-x\right)+x+x=P_i+x$ $x=P_t-P_i$ Here, $P_A=P_i-x \Rightarrow P_i-\left(P_t-P_i\right)$ $P_A=2 P_i-P_t$ $k=\frac{2.303}{t} \log \frac{P_i}{P_A}$ $k=\frac{2.303}{t} \log \frac{P_i}{\left(2 P_i-P_t\right)}$

Hence, the correct answer is option: 'B'.

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