Question

At 500K, the half-life period of a gaseous reaction at an initial pressure of 100kPa is 364s. When the pressure is 50kPa, the half-life period is 182s. The order of the reaction is.

• A. zero
• B. one
• C. half
• D. two

Answer 1

Answer: (A)

zero

Answer 2

For a gaseous reaction of order $n$, the half-life period ($t_{1/2}$) is related to the initial pressure ($P_0$) by the expression:

$t_{1/2} \propto \frac{1}{P_0^{n-1}}$

Considering two different initial pressures, $P_1$ and $P_2$, and their corresponding half-lives, $(t_{1/2})_1$ and $(t_{1/2})_2$, we can write the ratio:

$\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left(\frac{P_2}{P_1}\right)^{n-1}$

Given data:

$(t_{1/2})_1 = 364 \, \text{s}$ at $P_1 = 100 \, \text{kPa}$

$(t_{1/2})_2 = 182 \, \text{s}$ at $P_2 = 50 \, \text{kPa}$

Substitute these values into the equation:

$\frac{364}{182} = \left(\frac{50}{100}\right)^{n-1}$

$2 = \left(\frac{1}{2}\right)^{n-1}$

To solve for $n$, we can rewrite the equation using powers of 2:

$2^1 = (2^{-1})^{n-1}$

$2^1 = 2^{-(n-1)}$

$2^1 = 2^{-n+1}$

Equating the exponents:

$1 = -n + 1$

$1 - 1 = -n$

$0 = -n$

$n = 0$

The order of the reaction is 0.