Question

In the first order reaction, $A\left(g\right)\to B\left(g\right)+C\left(g\right),$ at constant volume and temperature, the initial pressure of $A$ is $11200$ $Pa$ and the total pressure at the end of $16$ minutes is $14667$ $Pa$. Calculate the half life period of reaction.

Answer 1

Ans : The reaction is

Given: $A\left(g\right)\to B\left(g\right)+C\left(g\right)$

Initial pressure of $A{P}_1=11200P_a$

Time $t=16\times 60=960sec$

Total Pressure after $P_2=14667P_a$

t time

Time $0$ $t$

Total Pressure $P_1$ $P_2$

Of $A+B+C$

The final pressure include the pressure of A,B and C.

But at t-0, the total pressure only include the

initial pressure of A. Let us assume same mdse

of B and C is formed by decom position of A

$A\left(g\right)\to B\left(g\right)+C\left(g\right)$

At $t=0$ $P_1$ $0$ $0$

At $t=0$ $P_1-x$ $x$ $x$

Let us conside x to be pressure of B and C at time t

Total pressure at time $t=P_1+x=P_2\Rightarrow x=P_2-P_1$

Now, the pressure of A at time +would be $=P_1-x$

$=P_1(P_2-P_1)=2P_1-P_2$

For first order reaction, relation between rate constant

and pressure is

$k=\frac{1}{t}ln\frac{P_1}{(2P_1-P_2)}$

$k=\frac{2.303}{t}log\frac{P_1}{2P_1-P_2}$

$2P_1-P_2-2\times 11260-14667\Rightarrow 22480-14667=77.33$

$k=\frac{2.303}{960}log\frac{11260}{7733}$

$k=\frac{2.303}{960}log1.44=3.74\times {10}^{-4}se{c}^{-1}$

Relation between half life and rate consider for

first order

$t_{t/2}=\frac{0.693}{k}$

$t_{1/2}=\frac{0.693}{3.79\times {10}^{-4}}=1.828\times {10}^{3}sec$