Question

A solution of $A$ is mixed with an equal volume of a solution of $B$ containing the same number of moles, and the reaction $A+B\to C$ occurs. At the end of $1h,A$ is $75$% reacted. How much of $A$ will be left unreacted at the end of $2h$ if the reaction is $\left(A\right)$ first order is $A$ and zero order in $B,\left(B\right)$ first order in both $A$ and $B$; and $\left(C\right)$ zero order in both $A$ and $B$?

Answer 1

The given reaction is :-

$A+B⟶C$

(A) The reaction is first order in $A$ and zero order in $B$

So, Rate=$K\left[A{]}^1\right[B{]}^0$

where,$K$ is rate constant

As the given reaction is of first order, so,

$K=\frac{1}{t}ln\frac{[A{]}_0}{[A{]}_t}$ $-\left(i\right)$

where,$K$= rate constant ,$t$=time

$[A{]}_0$=concentration of reactant at $t=0$

$[A{]}_t$=concentration of reactant at $t=t$

Now, at $t=1$ hour, $[A{]}_t=0.25[A{]}_0$

$(\because$ $75$% of $A$ is reacted, so amount of $A$ left unreacted is $25$%)

$\Rightarrow K=\frac{1}{1}ln\frac{\left[A_0\right]}{0.25\left[A_0\right]}$

$\Rightarrow K=ln4$ $hou{r}^{-1}$ $-\left(ii\right)$

Now, for a particular reaction $K$ will be constant.

So, after $t=2$ hours

$K=\frac{1}{2}ln\frac{[A{]}_0}{[A{]}_t}$

$\Rightarrow ln\frac{[A{]}_0}{[A{]}_t}=2\times ln4$ (from $\left(ii\right)$)

$\Rightarrow ln\frac{[A{]}_0}{[A{]}_t}=ln(4{)}^2$ $(\because ln$ $a^b=b$ $ln$ $a)$

$\Rightarrow \frac{[A{]}_0}{[A{]}_t}=4^2$

$\Rightarrow \frac{[A{]}_t}{[A{]}_0}=\frac{1}{16}$

so, percentage left=$\frac{[A{]}_t}{[A{]}_0}\times 100$%=$\frac{100}{16}$%$=6.25$%

(B) First order in both $A$ and $B$

So, Rate=$K\left[A\right]\left[B\right]$;$K$ is rate constant

Now, this reaction is of second order. But, it is of first order w.r.t $A$.

So, from the same treatment as done in $A$ we have,

% of $A$ left=$6.25$%

(C) zero order in both $A$ and $B$

So, Rate=$K\left[A{]}^0\right[B{]}^0$; $K$ is rate constant

Now, this reaction overall is a zero order reaction

so,

For zero order reaction we have:-

$K=\frac{1}{t}\left(\right[A{]}_0-\left[A{]}_t\right)$ $-\left(iii\right)$

where symbols have usual meaning as defined previously.

Now, $t=1$ hr,$[A{]}_t=0.25[A{]}_0$

From $\left(iii\right)$,

$K=\frac{1}{1}\left(\right[A{]}_0-0.25\left[A{]}_0\right)$

$\Rightarrow K=0.75[A{]}_0$

Now,after $t=2h$, we again have:-

$K=\frac{1}{t}\left(\right[A{]}_0-\left[A{]}_t\right)$

$\Rightarrow t=\frac{1}{0.75[A{]}_0}\times \left(\right[A{]}_0-\left[A{]}_t\right)$

$\Rightarrow 2\times 0.75=(1-\frac{[A{]}_t}{[A{]}_0})$

$\Rightarrow 1.5=1-\frac{[A{]}_t}{[A{]}_0}$

$\Rightarrow \frac{[A{]}_t}{[A{]}_0}=0.5$

So, percentage left= $50$%