# Problem 03-02-3

## Question

$$
2 \mathrm{X}(g)+2 \mathrm{Y}(g) \rightarrow \mathrm{Q}(g)+2 \mathrm{R}(g)
$$

The reaction represented above is found to be second order with respect to $\mathrm{X}$ and first order with respect to $\mathrm{Y}$.

What happens to the rate of the reaction when $[\mathrm{X}]$ is halved and $[\mathrm{Y}]$ is doubled?

-   [ ] It increases by a factor of 4.
-   [x] It decreases by a factor of 2.
-   [ ] It decreases by a factor of 4.
-   [ ] It does not change.

## Solution

To solve this problem, let's first write out the rate law for the reaction. According to the text, the reaction is second order with respect to $\mathrm{X}$ and first order with respect to $\mathrm{Y}$, so the rate law is rate $=k[\mathrm{X}]^2[\mathrm{Y}]$

Now, let's think about how the rate changes when $[\mathrm{X}]$ is halved and $[\mathrm{Y}]$ is doubled. Since $[\mathrm{X}]$ is raised to the second power in the rate law, halving $[\mathrm{X}]$ decreases the reaction rate by a factor of 4 . Similarly, since $[\mathrm{Y}]$ is raised to the first power, doubling $[\mathrm{Y}]$ increases the reaction rate by a factor of 2 . Combined, these changes result in the reaction rate decreasing by a factor of 2 overall.
So, when $[\mathrm{X}]$ is halved and $[\mathrm{Y}]$ is doubled, the rate of the reaction decreases by a factor of 2 .