Question

For the non-equilibrium process, $A+B\to Products$, the rate is first order with respect to $A$ and second-order with respect to $B$. If $1.0$ mole each of $A$ and $B$ are introduced into a 1-litre vessel and the initial rate was $1.0\times {10}^{-2}$ mol/litre-sec. The rate (in mol $litr{e}^{-1}se{c}^{-1}$) when half of the reactants have been used:

• A. $1.2\times {10}^{-3}$
• B. $1.2\times {10}^{-2}$
• C. $2.5\times {10}^{-4}$
• D. none of these

Answer 1

The correct option is A $1.2\times {10}^{-3}$

$rate=k\left[A\right]{\left[B\right]}^2$

Initially 1 mol each of A and B are present and the rate was $1\times {10}^{-2}mol/lit-sec$

When Half of the Reactants are used, the rate becomes $\frac{1}{2}\times {\left[\frac{1}{2}\right]}^2\times {10}^{-2}$ = $1.2\times {10}^{-3}$