Consider the reaction
3A + B + C $\rightarrow$ D + E
where the rate law is defined as k[A] 2 [B][C]
An experiment is carried out where [B] 0 = [C] 0 = 1.00 M and [A] 0 = 1.00 * 10 - 4 M.
-What is the concentration of A after 10.0 min?

Question

Consider the reaction
3A + B + C $\rightarrow$ D + E
where the rate law is defined as  k[A] 2 [B][C]
An experiment is carried out where [B] 0 = [C] 0 = 1.00 M and [A] 0 = 1.00 * 10 - 4 M.
-What is the concentration of A after 10.0 min?

Quizplus · Accepted Answer

We can use the integrated rate law for a 3rd order reaction, which is 1/[A] = kt + 1/[A]0 Plugging in the values from the experiment, we get 1/[A] = k(10.0 min) + 1/(1.00 × 10-6 M) 1/[A] = k(600 s) + 1.00 × 106 Now we need to find k, which can be done by rearranging the rate law and plugging in the values from the experiment: k = (rate) / ([A][B][C]) k = (change in D or E concentration) / (initial concentration of A × initial concentration of B × initial concentration of C) k = (0.200 M - 0 M) / (1.00 × 10-6 M × 1.00 M × 1.00 M) k = 2.00 × 107 M-2 s-1 Plugging in the values for k and t, we get 1/[A] = (2.00 × 107 M-2 s-1) (600 s) + 1.00 × 106 1/[A] = 1.20 × 1015 + 1.00 × 106 1/[A] = 1.20 × 1015. The concentration of A is therefore [A] = 8.33 × 10-16 M, which is equivalent to 1.27 × 10-10 F1F1F1S1F1F1F10 10^-⁵ M. Therefore, the answer is C.

Consider the Reaction 3A + B + C →\rightarrow→ D + E Where the Rate Law Is Defined as

Consider the Reaction
3A + B + C $\rightarrow$ D + E
Where the Rate Law Is Defined as