Question

For first order homogeneous gaseous reaction $A\to 2B+C$, the initial pressure was $P_0$ while total pressure at time '$t$' was $P_t$. The expression for the rate constant $k$ in terms of $P_0,P_t$ and $t$ is:

• A. $k=\frac{2.303}{t}log\left(\frac{2P_0}{2P_t-P_t}\right)$
• B. None of these
• C. $k=\frac{2.303}{t}log\left(\frac{2P_0}{3P_0-P_t}\right)$
• D. $k=\frac{2.303}{t}log\left(\frac{P_0}{P_0-P_t}\right)$

Answer 1

The correct option is C $k=\frac{2.303}{t}log\left(\frac{2P_0}{3P_0-P_t}\right)$

Progress of gaseous reaction can be monitored by measuring total pressure at a fixed volume and temperature.
We know,
$PV=nRT$
$V$ : Volume
$n$ : No. of moles
$P$ : Pressure
$R$ : Gas constant
$T$ : Temperature

So,
$P=\frac{n}{V}RTP\propto \frac{n}{V}P\propto \text{Concentration}$

So, for gaseous reactions, pressure is considered to monitor the reaction instead of concentration.

The first-order reaction is
$A\left(g\right)\to 2B\left(g\right)+C\left(g\right)$
$Att=0:P_{0}00$
$Att=t:P_0-P_{r}2P_r{P}_r$

$P_0$ is initial pressure
$P_r$ is pressure of reactant which is already converted to product

$\text{Total pressure,}{P}_t=(P_0-P_r)+2P_r+P_r$
$\Rightarrow P_t=P_0+2P_r$
$\Rightarrow P_r=\frac{1}{2}(P_t-P_0)$

So, pressure of $A$ after time $t$,
$P_A=P_0-P_r$
$\Rightarrow P_A=P_0-\frac{1}{2}(P_t-P_0)$
$\Rightarrow P_A=\frac{1}{2}(3P_0-P_t)$

Hene, the expression for the rate constant will be:
$k=\frac{2.303}{t}log\left(\frac{a}{a-x}\right)$

Since,
$P\propto \text{Concentration}$
$\frac{a}{a-x}$ = $\frac{P_0}{P_0-P_r}$

$\Rightarrow k=\frac{2.303}{t}log\left(\frac{P_0}{P_0-P_r}\right)$

Thus the expession for the rate constant becomes
$\Rightarrow k=\frac{2.303}{t}log\left(\frac{2P_0}{3P_0-P_t}\right)$
Hence, the correct option is A