$\begin{array}{1 1}(a)\;10minutes&(b)\;5minutes\\(c)\;20minutes&(d)\;40minutes\end{array}$

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Reaction is $A\rightarrow B+C$

Let initially there are no moles of A are present .

$A\rightarrow B+C$

At $t=0,A\Rightarrow n_0$

At $t=t,A\Rightarrow n_0-\lambda,B\Rightarrow \lambda,C\Rightarrow \lambda$

At $t=\infty,B\Rightarrow n_0,C\Rightarrow n_0$

As $\alpha=n[\alpha]$,at $t=\infty,\alpha_{obs}=-20^o=n_0(40)-n_0(80)$

We get $n_0=1/2$

Now $(\alpha_{\infty}-\alpha_t)\propto n_0-\lambda$ and $(\alpha_{\infty}-\alpha_0)\propto n_0$

We get $k=\large\frac{1}{t}$$ln\large\frac{n_0}{n_0-\lambda}=\large\frac{1}{t}$$ln\big(\large\frac{\alpha_{\infty}-\alpha_0}{\alpha_{\infty}-\alpha_t}\big)$

$\Rightarrow \large\frac{1}{t}$$ln\big(\large\frac{-20-30}{-20-5}\big)=\large\frac{ln\;2}{20}$$mol^{-1}$

$t_{1/2}=\large\frac{ln\;2}{K}$

$\;\;\;\;\;\;=20min$

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