Question

For the first order homogeneous gaseous reaction $A \to 2B + C$, the initial pressure was ${P_i}$ while total pressure at time $t$ was ${P_t}$. Write an expression for the rate constant $k$ in terms of ${P_i},{P_t}\& t$.
A. \[k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{2{P_i}}}{{3{P_i} - {P_t}}}} \right)\]
B. \[k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{2{P_i}}}{{2{P_t} - {P_i}}}} \right)\]
C. \[k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{P_i}}}{{{P_i} - {P_t}}}} \right)\]
D. None of these

Answer 1

Hint: A first order gaseous phase reaction involves only one reactant. Also, the rate of such a reaction is directly proportional to the amount of the reactant present. And in case of gas phase reaction partial pressure of the reactant is taken instead of the amount of reactant.

Formula Used:
\[k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{P_i}}}{{{P_A}}}} \right)\]
where ${P_A}$ is the partial pressure of the reactant at time $t$ and ${P_i}$ is the initial pressure of reactant $A$.

Complete step by step solution:
The given first order homogeneous equation is $A \to 2B + C$. Also, ${P_i}$ is the initial pressure of $A $ before the reaction.
Let $x$ denote the change in partial pressure at time $t$.
Then we can observe that the partial pressures of each gas are:

Time	A	B	C
$0$	${P_i}$	$0$	$0$
$t$	${P_i} - x$	$2x$	$x$

We can now use the expression for the rate constant and get,
$k = \dfrac{{2.303}}{t}\log \dfrac{{{P_i}}}{{{P_i} - x}}{\text{ - - - - - - - - - - - (1)}}$ \[\]
But note that,
 ${P_t} = {P_i} - x + 2x + x$
$ \Rightarrow {P_t} = {P_i} + 2x$
$ \Rightarrow x = \dfrac{1}{2}({P_t} - {P_i}){\text{ and }}{P_i} - x = \dfrac{1}{2}(3{P_i} - {P_t})$
Thus equation (1) becomes,
\[k = \dfrac{{2.303}}{t}\log \dfrac{{2{P_t}}}{{(3{P_i} - {P_t})}}\].
Hence, the answer is option A.

Note: Always note the number of molecules of a particular gas in the given equation before proceeding with calculations. If suppose one molecule of B is involved in the above reaction then its partial pressure at time $t$ would be just $x$ instead of $2x$ and the answer would be completely different.