The reaction
3C + D $\rightarrow$ products
Has rate = k[D] and has a half-life of 1.29 s.If initially [D] = 0.820,What is the concentration of D after 2.55 s?
A) 0.21
B) 0.12
C) 0.56
D) 0.28
E) 0.41

Question

Quizplus · Accepted Answer

The rate law is first order in D, so the half-life is independent of [D]. The half-life is given by t1/2 = ln(2)/k, so k = ln(2)/t1/2 = ln(2)/1.29 s = 0.539 s^-1. The integrated rate law is ln([D]/[D]0) = -kt, so [D] = [D]0e^(-kt) = 0.820e^(-0.539 s^-1 * 2.55 s) = 0.21 M.

The Reaction 3C + D →\rightarrow→ Products Has Rate = K[D] and Has a Half-Life of K[D]

The Reaction
3C + D $\rightarrow$ Products
Has Rate = K[D] and Has a Half-Life of K[D]