Question

For the following reaction, we have $-\frac{d\left[A\right]}{dt}=K\left[A{]}^2\right[B]$.

$2A+B\to C+D$

The expression for $-\frac{d\left[B\right]}{dt}$ would be:

• A. $K\left[A{]}^2\right[B]$
• B. $\frac{1}{2}K\left[A{]}^2\right[B]$
• C. $K\left[A{]}^2\right[2B]$
• D. $K\left[2A\right]\left[B\right]$

Answer 1

Answer: (B)

$\frac{1}{2}K\left[A{]}^2\right[B]$

For any reaction, the rate of reaction is given either by the change in the concentration of the reactants per unit time or the change in the concentration of the products per unit time.

Given a reaction:

$2A+B\to C+D$

For this, we have
$-\frac{1}{2}\frac{d\left[A\right]}{dt}$ $=$ $-\frac{d\left[B\right]}{dt}=\frac{d\left[C\right]}{dt}$ $=\frac{d\left[D\right]}{dt}$

So, from the above equation, we can say

$-\frac{1}{2}\frac{d\left[A\right]}{dt}$ $=$ $-\frac{d\left[B\right]}{dt}=k$

$-\frac{d\left[A\right]}{dt}=2k=K\left[A{]}^2\right[B]$ (given)

Now, $-\frac{d\left[B\right]}{dt}=k=\frac{1}{2}K\left[A{]}^2\right[B]$

Hence, the expression for $-\frac{d\left[B\right]}{dt}=\frac{1}{2}K\left[A{]}^2\right[B]$

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