Consider the following first order competating reactions:
X $k_{1}$ A + B and Y $k_{2}$ C + D
If 50% of the reaction of X was completed when 96% of the reaction of Y was completed, the ratio of their rate constants ( $k_{1} / k_{2}$ ) is: (Given $lo g 2 = 0.30$ , $lo g 3 = 0.48$ )

Question

Consider the following first order competating reactions:

X $k_{1}$ A + B and Y $k_{2}$ C + D

If 50% of the reaction of X was completed when 96% of the reaction of Y was completed, the ratio of their rate constants ( $k_{1} / k_{2}$ ) is: (Given $lo g 2 = 0.30$ , $lo g 3 = 0.48$ )

TestHub · Accepted Answer

For a first-order reaction: t = k 2.303 ​ lo g a − x a ​ For X (50% completed): t = k 1 ​ 0.693 ​ For Y (96% completed, 4% left): t = k 2 ​ 2.303 ​ lo g 4 100 ​ = k 2 ​ 2.303 ​ lo g 25 Since the time is the same: k 1 ​ 0.693 ​ = k 2 ​ 2.303 × 1.4 ​ k 2 ​ k 1 ​ ​ = 3.224 0.693 ​ = 0.215 But the ratio asked is: k 1 ​ k 2 ​ ​ = 4.60

Chemistry - Chemical Kinetics Question with Solution | TestHub

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