Question

At $500 \,K$ , the half-life period of a gaseous reaction at an initial pressure of $80\, kPa$ is $350 \,s$ . When the pressure is $40 \,kPa$ , the half-life period is $175 \,s$ . The order of the reaction is

• A. zero
• B. one
• C. two
• D. three

Answer 1

Answer: (A)

zero

Answer 2

The correct option is(A): zero.

${{P}_{1}}=80\,kPa,({{t}_{1/2}})=350s$
${{P}_{2}}=40\,kPa,{{({{t}_{1/2}})}_{2}}=175s$
$\frac{80}{40}=\frac{350}{175}=2$
$\because$ $\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{({{t}_{1/2}})}_{1}}}{{{({{t}_{1/2}})}_{2}}}=\frac{{{a}_{1}}}{{{a}_{2}}}$
$\therefore$ $={{t}_{1/2}}\propto a$ (zero order reaction)
Note: For ${{n}^{th}}$ order reaction ${{t}_{1/2}}\propto \frac{1}{{{(a)}^{n-1}}}$