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A reaction is first order with respective to A and second order with respective to $B$. What is the effect on reaction rate if concentration of B is increased 3 times?
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Verified Answer
The correct answer is:
Rate increase 9 times
Rate law of reaction,
$$
\mathrm{r}=\mathrm{k}[\mathrm{A}]^{\prime}[\mathrm{B}]^2
$$
If concentration of $\mathrm{B}$ is increased 3 times,
$$
\begin{aligned}
& \mathrm{B} \rightarrow 3 \mathrm{~B} \\
& \mathrm{r}_{\mathrm{New}}=\mathrm{k}[\mathrm{A}][3 \mathrm{~B}]^2 \\
& \frac{\mathrm{r}_{\mathrm{New}}}{\mathrm{r}}=\frac{\mathrm{k}[\mathrm{A}][3 \mathrm{~B}]^2}{\mathrm{k}[\mathrm{A}][\mathrm{B}]^2} \\
& \frac{\mathrm{r}_{\mathrm{New}}}{\mathrm{r}}=3^2 \\
& \mathrm{r}_{\mathrm{New}}=9 \mathrm{r}
\end{aligned}
$$
Rate increase 9 times.
$$
\mathrm{r}=\mathrm{k}[\mathrm{A}]^{\prime}[\mathrm{B}]^2
$$
If concentration of $\mathrm{B}$ is increased 3 times,
$$
\begin{aligned}
& \mathrm{B} \rightarrow 3 \mathrm{~B} \\
& \mathrm{r}_{\mathrm{New}}=\mathrm{k}[\mathrm{A}][3 \mathrm{~B}]^2 \\
& \frac{\mathrm{r}_{\mathrm{New}}}{\mathrm{r}}=\frac{\mathrm{k}[\mathrm{A}][3 \mathrm{~B}]^2}{\mathrm{k}[\mathrm{A}][\mathrm{B}]^2} \\
& \frac{\mathrm{r}_{\mathrm{New}}}{\mathrm{r}}=3^2 \\
& \mathrm{r}_{\mathrm{New}}=9 \mathrm{r}
\end{aligned}
$$
Rate increase 9 times.
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