For the irreversible process , the rate is first-order with respect to and second-order with respect — Chemical Kinetics Chemistry Question
Question
For the irreversible process $A + B \rightarrow \text{Products}$, the rate is first-order with respect to $A$ and second-order with respect to $B$. If $1.0 \text{ mol}$ each of $A$ and $B$ are introduced into a $1.0 \text{ L}$ vessel, and the initial rate was $1.0 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}$, what is the rate when half of the reactants have been turned into products?
💡 Solution & Explanation
Rate $= k[A][B]^2$. Initially, $1.0 \times 10^{-2} = k(1.0)(1.0)^2 \Rightarrow k = 1.0 \times 10^{-2} \text{ L}^2 \text{ mol}^{-2} \text{ s}^{-1}$. When half the reactants are consumed, $[A] = 0.5 \text{ M}$ and $[B] = 0.5 \text{ M}$. New Rate $= (1.0 \times 10^{-2}) \times (0.5)^1 \times (0.5)^2 = 1.25 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}$.