# Problem 03-02-3 | Toshiki's Note
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# Problem 03-02-3 ​

Author:Anda Toshiki
Updated:a minute ago
Words:246
Reading:1 min

## Question ​

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

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

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

- It increases by a factor of 4.
- 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 X\mathrm{X}X and first order with respect to Y\mathrm{Y}Y, so the rate law is rate =k[X]2[Y]=k[\mathrm{X}]^2[\mathrm{Y}]=k[X]2[Y]

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