Question

For a chemical reaction, $2A+2B\to C+D$ , the order of reaction is one with respect to A and one with respect to B. The initial rate of the reaction is $4\times {10}^{-2}mol$ $L^{-1}{s}^{-1}$ . When 50% of the reactant are converted into products, the rate of the reaction would become:

• A. $2\times {10}^{-2}mol$ $L^{-1}{s}^{-1}$
• B. $1\times {10}^{-2}mol$ $L^{-1}{s}^{-1}$
• C. $4\times {10}^{-2}mol$ $L^{-1}{s}^{-1}$
• D. $2\times {10}^{-1}mol$ $L^{-1}{s}^{-1}$

Answer 1

The correct option is B $1\times {10}^{-2}mol$ $L^{-1}{s}^{-1}$

$rate=k\left[A\right]\left[B\right]$

$4\times {10}^{-2}=k\left[A\right]\left[B\right]$

Now,

$\left[A\right]\to \frac{\left[A\right]}{2}$ $\left[B\right]\to \frac{\left[B\right]}{2}$

$(\text{rate}{)}_{new}=k\frac{\left[A\right]}{2}\frac{\left[B\right]}{2}$

$=\frac{4\times {10}^{-2}}{4}$

$=1\times {10}^{-2}$

Hence, the correct option is $\text{B}$