Question

For first order homogeneous gaseous reaction A → 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 $P_0,P_t$ & t is:

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

Answer 1

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

Progress of gaseous reaction can be monitored by measuring total pressure at a fixed volume & 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

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

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

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

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

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

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