Consider the reaction $A_{2} + B \to p r o d u c t s$ . If the concentration of $A_{2}$ and B are halved, the rate of the reaction decreases by a factor of 8. If the concentration of A is increased by a factor of 2.5, the rate increases by the factor of 2.5. What is the order of the reaction? Write the rate law.

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Solution

Let the rate of reaction be
$R a t e (1) = k {[A_{2}]}^{x} {[B]}^{y}$
When concentration of [A] $&$ [B] are
halved then new rate is
Rate $(2) = k {[\frac{A_{2}}{2}]}^{x} {[\frac{B}{2}]}^{y}$
It is given that $8 \times R a t e (2) = R a t e (1) ∴ \frac{R a t e (2)}{R a t e (1)} = \frac{1}{8} = \frac{k {[\frac{A_{2}}{2}]}^{x} {[\frac{B}{2}]}^{y}}{k {[A_{2}]}^{x} {[B]}^{y}} {(\frac{1}{2})}^{3} = \frac{1}{8} = {(\frac{1}{2})}^{x + y} ∴ x + y = 3 ⟶ (1)$
It is given that $2.5 \times R a t e (1) = R a t e (3) \frac{R a t e (3)}{R a t e (1)} = \frac{k {[{2.5 A}_{2}]}^{x} {[B]}^{y}}{k {[A_{2}]}^{x} {[B]}^{y}} = 2.5 {(2.5)}^{x} = 2.5 ∴ x = 1, \Rightarrow y = 2$
A is following first order, B-is following second order, and reaction is third order.

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Consider the reaction A2+B→products. If the concentration of A2 and B are halved, the rate of the reaction decreases by a factor of 8. If the concentration of A is increased by a factor of 2.5, the rate increases by the factor of 2.5. What is the order of the reaction? Write the rate law.