Question

the reaction a +2b $
ightarrow$ products has the rate law, rate= $kab^3$. if the concentration of b is doubled while that of a is unchanged, by what factor will the rate of reaction increase?
8
6
9
4
2

Explanation:

Step1: Define initial reaction rate

$\text{rate}_1 = k[\text{A}][\text{B}]^3$

Step2: Substitute doubled [B]

$\text{rate}_2 = k[\text{A}](2[\text{B}])^3$

Step3: Simplify the new rate

$\text{rate}_2 = k[\text{A}] \times 8[\text{B}]^3 = 8k[\text{A}][\text{B}]^3$

Step4: Find rate ratio

$\frac{\text{rate}_2}{\text{rate}_1} = \frac{8k[\text{A}][\text{B}]^3}{k[\text{A}][\text{B}]^3} = 8$

Answer:

8