Chemical Kinetics Question 9
Question 9 - 2024 (31 Jan Shift 1)
Integrated rate law equation for a first order gas phase reaction is given by (where $P_{i}$ is initial pressure and $\mathrm{P}_{\mathrm{t}}$ is total pressure at time $\mathrm{t}$ )
(1) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{\mathrm{P}{\mathrm{i}}}{\left(2 \mathrm{P}{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}$
(2) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{2 \mathrm{P}{\mathrm{i}}}{\left(2 \mathrm{P}{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}$
(3) $\mathrm{k}=\frac{2.303}{\mathrm{t}} \times \log \frac{\left(2 \mathrm{P}{\mathrm{i}}-\mathrm{P}{\mathrm{t}}\right)}{\mathrm{P}_{\mathrm{i}}}$
(4) $k=\frac{2.303}{t} \times \frac{P_{i}}{\left(2 P_{i}-P_{t}\right)}$
Show Answer
Answer (1)
Solution
$\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}$
$\begin{array}{lll}P_{i} & 0 & 0\end{array}$
$P_{i}-x$ x $x$
$P_{t}=P_{i}+x$
$P_{i}-x=P_{i}-P_{t}+P_{i}$
$=2 P_{i}-P_{t}$
$K=\frac{2.303}{t} \log \frac{P_{i}}{2 P_{i}-P_{t}}$
