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  • The half life periods of a reaction at initial concentraions 0.1 mol L-1 and 0.5 mol L-1 are 200 s and 40 s, respectively. The order of the reaction is:<div clas

The half life periods of a reaction at initial concentraions 0.1 mol L-1 and 0.5 mol L-1 are 200 s and 40 s, respectively. The order of the reaction is:

Option: 1

1/2


Option: 2

3/2


Option: 3

1


Option: 4

2


Answers (1)

best_answer

How to Determine Order of Reaction: Half-Life Method -

It is used when the rate law involves only one concentration term.

\mathrm{t_{1 / 2} \propto(a)^{1-n}} 
 

For two different concentrations, we have:

\mathrm{\frac{\left(t^{1 / 2}\right)_{1}}{\left(t^{1 / 2}\right)_{2}}=\left(\frac{a_{2}}{a_{1}}\right)^{n-1}}

Now,

The half-life periods of a reaction at initial concentrations 0.1 mol L-1 and 0.5 mol L-1 are 200 s and 40 s, respectively. 

\begin{array}{l}{[\mathrm{A}]_{0}=0.1 \mathrm{M} \longrightarrow t_{1 / 2}=200 \mathrm{\: s}\: \: \: \: \: ....(\mathrm i)} \\ {[\mathrm{A}]_{0}=0.5 \mathrm{M} \longrightarrow t_{1 / 2}=40 \mathrm{\: s}\: \: \: \: \: ....(\mathrm{ii})} \end{array}

From cases (i) and (ii), we can conclude 
\mathrm{\frac{200}{40}=\left(\frac{0.5}{0.1}\right)^{n-1}}

\mathrm{\left (\frac{5}{1} \right )^1=\left(\frac{5}{1}\right)^{n-1}}

So, 1 = n - 1

then, n = 2

Therefore, it is a second-order reaction.

Therefore, option(4) is correct

Posted by

Gaurav

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