‹ Back to the paper
Answer the following.
In the reaction , the rate law is rate . How will the rate of reaction change if concentration is…
In the reaction $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$, the rate law is rate $= k[\text{NO}][\text{O}_2]^2$.
How will the rate of reaction change if $[\text{NO}]$ concentration is doubled and $[\text{O}_2]$ concentration is halved at the same time?
Answer
Answer
AIThe rate becomes half (it decreases by 50%).
Initially rate $= k[\text{NO}][\text{O}_2]^2$. New rate $= k(2[\text{NO}])\left(\dfrac{[\text{O}_2]}{2}\right)^2 = \dfrac{1}{2}\,k[\text{NO}][\text{O}_2]^2$.
From ISC 2025 Chemistry Paper 1, question 4(i).