Question

Which of the following represents the rate law for the following reversible reaction?

• A. $-\frac{d\left[A\right]}{dt}=k_1\left[A\right]+k_2\left[B\right]$
• B. $\frac{+d\left[A\right]}{dt}=k_1\left[A\right]-k_2\left[B\right]$
• C. $-\frac{d\left[B\right]}{dt}=k_1\left[A\right]-k_2\left[B\right]$
• D. $-\frac{d\left[B\right]}{dt}=k_1\left[A\right]-k_2\left[B\right]$

Answer 1

The correct option is C $-\frac{d\left[B\right]}{dt}=k_1\left[A\right]-k_2\left[B\right]$

Let us write down the rate law for the reaction using what we know.

We know that rate of reduction in [A] and rate of increase in [B] are equal. So,

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

And we know that this is proportional [A]. But B is produced as the reaction starts and also starts converting back to A at the given rate $k_2$

Therefore, we need to reverse sign of [B] in the rate law to account for this change.

We get,

$-\frac{d\left[A\right]}{dt}=\frac{d\left[B\right]}{dt}=k_1\left[A\right]-k_2\left[B\right]$