For the second-order reaction below,the concentration of product B after 132 seconds is 0.0281 M.If the initial concentration of A is 0.932 M,what is the rate constant? $2 \mathrm{~A} \rightarrow \mathrm{B} \quad\quad\quad\quad\quad \text { rate }=k[\mathrm{~A}]^{2}$

Question

Quizplus · Accepted Answer

The concentration of A at any time t can be calculated using the following equation:

$$[\mathrm{A}]=\frac{1}{\sqrt{\frac{1}{[\mathrm{A}]_{0}^{2}}+4 k t}}$$

where $[\mathrm{A}]_{0}$ is the initial concentration of A.

Using the given values, we can solve for the concentration of A at t = 132 s:

$$[\mathrm{A}]=\frac{1}{\sqrt{\frac{1}{(0.932 \mathrm{~M})^{2}}+4 k (132 \mathrm{~s})}}=0.358 \mathrm{~M}$$

Now, we can use the concentration of A and B at t = 132 s to solve for the rate constant k:

$$k=\frac{	ext {rate}}{[\mathrm{A}]^{2}}=\frac{\frac{\Delta [\mathrm{B}]}{\Delta t}}{[\mathrm{A}]^{2}}=\frac{(0.0281 \mathrm{~M}-0.0) / \mathrm{~s}}{(0.358 \mathrm{~M})^{2}}=5.22 	imes 10^{-4} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}$$

For the Second-Order Reaction Below,the Concentration of Product B After 2 A→B rate =k[ A]22 \mathrm{~A} \rightarrow \mathrm{B} \quad\quad\quad\quad\quad \text { rate }=k[\mathrm{~A}]^{2}2 A→B rate =k[ A]2

For the Second-Order Reaction Below,the Concentration of Product B After $2 \mathrm{~A} \rightarrow \mathrm{B} \quad\quad\quad\quad\quad \text { rate }=k[\mathrm{~A}]^{2}$