Consider the reaction 3A + B + C $\to$ D + E where the rate law is defined as .
An experiment is carried out where [B] 0 = [C] 0 = 1.00 M and [A] 0 = 1.00 $\times$ 10 - 4 M.
-The concentration of A after 10.0 minutes is

Question

Consider the reaction 3A + B + C $\to$ D + E where the rate law is defined as  .
An experiment is carried out where [B] 0 = [C] 0 = 1.00 M and [A] 0 = 1.00 $\times$ 10 - 4 M.
-The concentration of A after 10.0 minutes is

Quizplus · Accepted Answer

The rate law is defined as 11eaa8f0_66fa_5338_93a6_f9c169f1857b_TB6423_11_TB6423_11_TB6423_11_TB6423_11 . As we see that the concentration of B and C remains the same throughout the reaction, we can consider their concentration as constant. Hence, the rate law can be simplified as follows: rate = k'[A] Integrating it gives: ln([A]S1U1F1F1S1F1F1F10 10^-⁴ / [A]₀) = -k't Substituting the given values, we get: ln(1.06) = -k'(10.0) Solving for k' gives: k' = 0.076 Now, using this value of k' in the rate law, we can calculate the concentration of A after 10.0 minutes as follows: rate = 0.076[A] At t = 0, [A] = 1.00 F1F1F1S1F1F1F10 10^-⁴ M At t = 10.0 min, let [A] = x. Then, 0.076x = ln(1.06) x = 1.27 F1F1F1S1F1F1F10 10^-⁵ M Hence, the correct answer is D.

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

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