Question

The rate for reaction A + B $\rightarrow$ product, is $1.8\times10^{-2}\text{ mol dm}^{-3}\text{s}^{-1}$. Calculate the rate constant if the reaction is second order in A and first order in B. $([A]=0.2\text{ M}; [B]=0.1\text{ M})$

• A. $9.0\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$
• B. $18.0\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$
• C. $4.5\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$
• D. $16.0\text{ mol}^{-2}\text{dm}^{6}\text{s}^{-1}$

Hint:

Logic Tip: To make the division easier without using scientific notation, convert decimals to fractions: $0.2 = 2/10$ and $0.1 = 1/10$. Then $Rate = k \times (4/100) \times (1/10) = k \times (4/1000)$. So, $k = 0.018 \times (1000/4) = 18/4 = 4.5$.

Answer 1

The Correct Option is C