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

Hint: 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.