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Consider the reaction A + B ? products
From the following data obtained at a certain temperature, determine the rate law, the order of the reaction, and calculate the rate constant k.
Experiment 1: [A] = 1.50 M; [B] = 1.50 M; Initial Rate = 3.20 x 10-1 M/s
Experiment 2: [A] = 1.50 M; [B] = 2.50 M; Initial Rate = 3.20 x 10-1 M/s
Experiment 3: [A] = 3.00 M; [B] = 1.50 M; Initial Rate = 6.40 x 10-1 M/s
Please explain to me how you got the answer step by step >< Thank you!
A. Rate = k[A][B]
order of reaction = 2
k = 0.142 s-1
B. Rate = k[A]
order of reaction = 1
k = 0.213 s-1
C. Rate = k[A]2
order of reaction = 2
k = 0.142 s-1
D. Rate = k[B]
order of reaction = 1
k = 0.213 s-1

Respuesta :

sebassandin

Answer:

B. Rate = k[A]

order of reaction = 1

k = 0.213 s-1

Explanation:

Hello,

In this case, you should consider the experiments and how the concentration of both A and B affect the rate, thus, we can make some conclusions:

- Between the experiment 1 and 2, we can notice that modifying the concentration of B does not affect the rate as it remains in 3.20 x 10-1 M/s, for that reason the reaction is zeroth-order respect to B.

- Between the experiment 2 and 3, we can notice that doubling the concentration of A from 1.50 M to 3.00 M results in a doubling of the rate from 3.20 x 10-1 M/s to 6.40 x 10-1 M/s, for that reason, we can infer that the reaction is first-order respect to A.

In such a way, we can infer that the rate law is:

[tex]Rate=k[A]^1[B]^0\\\\Rate= k[A][/tex]

Thus, the order of reaction is first-order and the rate constant turns out:

[tex]k=\frac{Rate}{[A]}=\frac{3.2x10^{-1}M/s}{1.50M}\\ \\k=0.213s^{-1}[/tex]

Therefore, answer is:

B. Rate = k[A]

order of reaction = 1

k = 0.213 s-1

Regards.

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