Question

For a hypothetical elementary reaction wheree $\frac{k_1}{k_2}=\frac{1}{2}$. Initially only $2$ moles of ${}^{\prime }{A}^{\prime }$ is present. Then find the total number of moles $A,B$ and $C$ at the end of $50$% reaction.

Answer 1

$\frac{d\left[A\right]}{dt}=(k_1+k_2)$

$-d\left[A\right]=(k_1+k_2)dt$

$-\left[A\right]=(k_1+k_2)t+c$

at $t=0$

$-\left[A_0\right]=c\Rightarrow -\left[A\right]=(k_1+k_2)t+[-[A_0\left]\right]$

$\frac{\left[A_0\right]-\left[A\right]}{k_1+k_2}=t$

After 50% completion $\left[A\right]=1mole$

$t=\frac{2-1}{3}=\frac{1}{3}$

$\left[B\right]=k_{1}t$

$=\frac{1}{3}$ mole

$\left[C\right]=k_{2}t$

$=\frac{2}{3}mole$