Question

What will be the order of reaction and rate constant for a chemical change having $\log t_{50}$% $vs$ $\log$ concentration of $\left(A\right)$ curves as:

• A. $0,1/2$
• B. $1,1$
• C. $2,2$
• D. $3,1$

Answer 1

The correct option is A $0,1/2$

We know that,

Half time of $n^{th}$ order reaction $\left(t_{1/2}\right)\propto {\left(\frac{1}{[A{]}_0}\right)}^{n-1}$

where $[A{]}_0$ is the initial concentration of the reactant

So, $t_{1/2}=\frac{C}{\left(\right[A{]}_0{)}^{n-1}}$ $-\left(i\right)$; $C$ is proportionality constant.

Taking $\log$ on both sides of $\left(i\right)$:-

$\log t_{1/2}=\log C(n-1)\log \left[A\right]$

$[\because \log \left(\frac{a}{b}\right)=\log a-\log b$ & $\log \left(a^b\right)=b\log a]$

Now, for $n=0$,

$\log t_{1/2}=\log C-(0-1)log[A{]}_0$

$\Rightarrow log{t}_{1/2}=\log C+log[A{]}_0$

If $C=)$, then,

$\Rightarrow \log t_{1/2}=\log [A{]}_0$ $-\left(ii\right)$

$E{q}^n\left(ii\right)$ represents the graph between $\log t_{1/2}$ & $\log [A{]}_0$ as the straight line passing through origin with an angle of ${45}^{\circ }$ from $\log [A{]}_0$ axis.

(Similarly as line $y=x$)

Now, for a zero order reaction,

$t_{1/2}=\frac{[A{]}_0}{2K}$

where, $K$ is the rate constant.

$\Rightarrow K=\frac{[A{]}_0}{2t_{1/2}}$ $-\left(iii\right)$

From the graph we have:-

$\log t_{1/2}=\log [A{]}_0$

$\Rightarrow t_{1/2}=[A{]}_0$ (Taking antilog on both sides)

So, from $\left(iii\right)$, $K=\frac{[A{]}_0}{2[A{]}_0}=\frac{1}{2}$