Question

For a reaction: $ A + B \rightarrow \text{Products} $ It is found that doubling the concentration of A doubles the rate, while doubling B increases the rate fourfold. What is the overall order of the reaction?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Use the rate law equation and the given data to determine the individual orders of reactants, then sum them to find the overall order of the reaction.

Answer 1

The Correct Option is C

The rate law for the reaction can be written as: \[ \text{Rate} = k [A]^m [B]^n \] where: - \(m\) is the order of the reaction with respect to \(A\), - \(n\) is the order of the reaction with respect to \(B\), - \(k\) is the rate constant. Doubling the concentration of A: - When the concentration of \(A\) is doubled, the rate doubles. This implies that \(m = 1\) because: \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k [2A]^m [B]^n}{k [A]^m [B]^n} = 2 \quad \Rightarrow \quad 2^m = 2 \quad \Rightarrow \quad m = 1 \] Doubling the concentration of B: - When the concentration of \(B\) is doubled, the rate increases fourfold. This implies that \(n = 2\) because: \[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k [A]^m [2B]^n}{k [A]^m [B]^n} = 4 \quad \Rightarrow \quad 2^n = 4 \quad \Rightarrow \quad n = 2 \] Thus, the overall order of the reaction is \(m + n = 1 + 2 = 3\).
Final answer
Answer: \(\boxed{3}\)