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The thermal decomposition of phosphine to phosphorus and hydrogen gas is a first order reaction, with a half-life of 35 s at 680$^\circ$C. Calculate the time required for 95% of phosphine to decompose.

$\begin{array}{1 1} 151\;s \\ 302\;s \\ 255\;s \\ 454\;s \end{array}$

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Answer: 151 s
Step-1: Calculate $k$, given $t_{1/2}$:
The half life $t_{1/2} = \large\frac{ ln\;2}{k}$$ \rightarrow k = \large\frac{0.693}{t_{1/2}}$$ = \large\frac{0.693}{35}$$ = 0.0198 s^{-1}$
Step-2: Calculate the time required for 95% of phosphine to decompose
We know that $\log \large\frac{[A_t]}{[A_0]}$$ = - \large\frac{kt}{2.303}$
$\Rightarrow t = - \log \large\frac{0.5}{1}$$\times \large\frac{2.303}{0.0198}$$ = -1.3\times -\large\frac{ 2.303}{0.0198}$ $=151\;s$
answered Jul 24, 2014 by balaji.thirumalai
 

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