At 300 ~K , the half-life period of a gaseous reaction at an initial pressure of 40 ~kPa is 350 s. When pressure is 20 ~kPa , the half-life period is 175 s. What is the order of the reaction?

Step-by-step solution

For a gaseous reaction of order n , the half-life t_ 1/2 is related to the initial pressure P₀ by the relation t_ 1/2 1 P₀^ n-1 . Given data: Case 1: P_ 0,1 = 40 kPa , t_ 1/2,1 = 350 s Case 2: P_ 0,2 = 20 kPa , t_ 1/2,2 = 175 s Taking the ratio of the two cases: t_ 1/2,1 t_ 1/2,2 = ( P_ 0,2 P_ 0,1 )^ n-1 Substituting the values: 350 175 = ( 20 40 )^ n-1 2 = ( 1 2 )^ n-1 2 = (2⁻¹)^ n-1 2^1 = 2^ 1-n Equating the exponents: 1 = 1 - n n = 0 . Answer: Zero