Solutions of two compounds, W and X, react together in the presence of a soluble catalyst, Y, as shown in the equation $$2W + X \rightarrow Z$$ When the concentrations of W, X and Y are all doubled, the rate of reaction increases by a factor of four - AQA - A-Level Chemistry - Question 12 - 2018 - Paper 3

Given that the concentrations of W, X, and Y are all doubled:

• New concentration of W: $[W]^{*} = 2[W]$
• New concentration of X: $[X]^{*} = 2[X]$
• New concentration of Y: $[Y]^{*} = 2[Y]$

We observe that the rate of reaction increases by a factor of four, thus:

$\text{rate}^{*} = k(2[W])^m(2[X])^n(2[Y])^p$

This simplifies to:

$\text{rate}^{*} = k \cdot 2^m[W]^m \cdot 2^n[X]^n \cdot 2^p[Y]^p = k \cdot 2^{m+n+p}[W]^m[X]^n[Y]^p$

Since the rate increases by a factor of four, we have:

$\text{rate}^{*} = 4 \cdot \text{rate} = k \cdot 2^{m+n+p}[W]^m[X]^n[Y]^p$

Setting the expressions equal involving the factor increase gives us:

$2^{m+n+p} = 4$

This means:

$m+n+p = 2$

This implies that the orders of reaction with respect to W, X, and Y combined equals 2.

Considering the impact of W and X on the rate equation while recognizing that Y is constant during initial analysis, we can examine the options:

• A: $\text{rate} = k[W]^2[X]$ (Sum is 2)
• B: $\text{rate} = k[W]^2[Y]$ (Invalid as Y must vary)
• C: $\text{rate} = k[W][X]$ (Sum is 2)
• D: $\text{rate} = k[X][Z]$ (Invalid as Z is a product)

Based on the calculations above, both A and C are possible, but since we focus on the contributions of W and X alone, the most suitable option that fits our derived requirement is:

C: $\text{rate} = k[W][X]$, which represents a valid rate expression under the given scenario.