Answer to Question #110471 in Organic Chemistry for Myra

Question

For a second order reaction, if 46% of the original 0.550M sample has decomposed after 45 seconds. What is the rate constant for the reaction?

Answer 1

Solution:

For 2^nd order reaction:

d[A]/dt = −k[A]²

Integration of the second-order rate law yields:

1/[A] =1/[A]₀ + kt

[A]^-1 = [A]₀^-1 + kt

Since 46% has decomposed, 54% remains. Therefore,

[A] = 0.54*[A]₀ = 0.54 * 0.550 M= 0.297 M.

Then,

[A]^-1 = [A]₀^-1 + kt

[0.297]^-1 = [0.550]^-1 + k*45;

3.367 = 1.818 + k*45;

[1.549 M^-1]= [45 s]*[k]

k = 0.0344 M^-1 s^-1.

Answer: 0.0344 M^-1 s^-1 is the rate constant (k) for the reaction.