For first order kinetics reaction,

A → B + C (3)

If initial pressure of A is 1 bar and at time 100 s , the total pressure is 1.5 bar, then find the rate constant of the reaction.

The correct answer is 0.00693 s^–1.

Step 1: Understand the reaction and initial conditions
Consider the given first-order gas-phase reaction:
$\text{A}\to \text{B}+\text{C}$
Let $P_0$ be the initial pressure of reactant A, and $x$ be the decrease in pressure of A due to reaction progress at time $t=100\text{s}$.

Step 2: Express partial pressures at time t = 0 and t = 100 s
At time $t=0$:
Partial pressure of reactant A, $P_{\text{A}}=P_0=1\text{bar}$
Partial pressure of product B, $P_{\text{B}}=0$
Partial pressure of product C, $P_{\text{C}}=0$

At time $t=100\text{s}$:
Partial pressure of A, $P_{\text{A}}=1-x$
Partial pressure of B, $P_{\text{B}}=x$
Partial pressure of C, $P_{\text{C}}=x$

Step 3: Calculate x using total pressure
The total pressure at $t=100\text{s}$ is given as 1.5 bar:
$P_{\text{total}}=P_{\text{A}}+P_{\text{B}}+P_{\text{C}}$
$1.5=(1-x)+x+x$
$1.5=1+x$
$x=1.5-1=0.5\text{bar}$

Step 4: Find the remaining pressure of A at t = 100 s
$P_{\text{A}}=1-0.5=0.5\text{bar}$

Step 5: Calculate the rate constant (k)
For a first-order reaction, the rate constant equation is given by:
$k=\frac{1}{t}\ln \left(\frac{P_0}{P_{\text{A}}}\right)$

Substitute the values into the formula:
$k=\frac{1}{100}\ln \left(\frac{1}{0.5}\right)$
$k=\frac{1}{100}\ln \left(2\right)$

Using the value of $\ln \left(2\right)\approx 0.693$:
$k=\frac{0.693}{100}=0.00693{\text{s}}^{-1}$

Therefore, the rate constant of the reaction is 0.00693 s^–1.