Question

For a reaction $A\stackrel{K_1}{⟶}B\stackrel{K_2}{⟶}C.$ If the reaction are of $1^{st}$ order then $\frac{d\left[B\right]}{dt}$is equal to:

• A. $-k\left[B\right]$
• B. $+k_1\left[A\right]$
• C. $k_1\left[A\right]-k_2\left[B\right]$
• D. $k_1\left[A\right]+k_2\left[B\right]$

Answer 1

Answer: (C)

$k_1\left[A\right]-k_2\left[B\right]$

For the first reaction $A\to B$
$\frac{d\left[B\right]}{dt}=k_{1}A$.....(1)
For the second reaction $B\to C$
$-\frac{d\left[B\right]}{dt}=k_2\left[B\right]$
or
$\frac{d\left[B\right]}{dt}=-k_2\left[B\right]$......(2)
Total $\frac{d\left[B\right]}{dt}$ can be obtained by adding equations (1) and (2)
$\frac{d\left[B\right]}{dt}=k_1\left[A\right]-k_2\left[B\right]$