Question

At $500\;K$ , the half-life period of a gaseous reaction at an initial pressure of $80\;kPa$ is $350\;\sec$ . When the pressure is $40\;kPa$ , the half-life period is $175\;\sec$ . the order of the reaction is:
(A) Zero
(B) One
(C) Two
(D) Three

Answer 1

The half-life period of a chemical reaction can be defined as the time taken by the reactant to have its concentration become half of its initial concentration or the time taken by concentration of a given reactant to reach half of its starting value. It is represented by ${{\text{t}}_{\dfrac{1}{2}}}$ , It is expressed in seconds. Here the half-life period is described in terms of pressure, we will calculate the order of the reaction by using the half-life of the given gaseous reaction.

Complete answer:
Let’s understand the order of reaction. suppose a simple reaction
$xX + yY \to zZ$
Rate law equation for above reaction is given as:
$Rate = {\text{K}}{\left[ {\text{X}} \right]^{\text{a}}}{\left[ {\text{Y}} \right]^{\text{b}}}$
where ${\text{a}} + {\text{b}}$ is the order of the reaction and ${\text{K}}$ is the Specific Rate constant X is the concentration of reactant X and Y is the concentration of reactant Y.
We are given,
Temperature for the reaction, ${\text{T}} = 500\;{\text{K}}$
Initial Pressure of the reaction, ${{\text{P}}_1} = \;80\;kPa$
First half-life period of the reaction, ${\left( {{{\text{t}}_{\dfrac{1}{2}}}} \right)_1} = 350\;sec$
final pressure of the reaction, ${{\text{P}}_2} = \;40\;kPa$
second half-life period of the reaction, ${\left( {{{\text{t}}_{\dfrac{1}{2}}}} \right)_2} = 175\;sec$
Putting it in Half-life formula, for now we assume n to be order of the reaction ,we get
${\left( {{{\text{t}}_{\dfrac{1}{2}}}} \right)_1} = \dfrac{1}{{K{{\left[ {{P_1}} \right]}^n}}}$
similarly for second half-life period of the reaction,
${\left( {{{\text{t}}_{\dfrac{1}{2}}}} \right)_2} = \dfrac{1}{{K{{\left[ {{P_2}} \right]}^n}}}$
dividing first half-life by second half-life, we get
$\dfrac{{{{\left( {{{\text{t}}_{\dfrac{1}{2}}}} \right)}_1}}}{{{{\left( {{{\text{t}}_{\dfrac{1}{2}}}} \right)}_2}}} = \dfrac{{{{\left[ {{P_2}} \right]}^{n - 1}}}}{{{{\left[ {{P_1}} \right]}^{n - 1}}}}$
substituting values in the above equation,
$\dfrac{{350}}{{175}} = {\left[ {\dfrac{{40}}{{80}}} \right]^{n - 1}}$
we get,
$2 = {\left[ {\dfrac{1}{2}} \right]^{n - 1}}$
$1 - {\text{n}} = 1$
so, ${\text{n}} = 0$ is the order of the reaction.
Hence, option (A) is the correct answer.

Note:
Order of a reaction is defined as the sum of powers of concentration of reactant and products in the Rate law equation. The order of a reaction represents the number of entities that are affecting the rate of the reaction.