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  • Help me solve this A reaction is 50% completed in 2 hours and 75% completed in 4 hours. The order of reaction is

A reaction is 50% completed in 2 hours and 75% completed in 4 hours. The order of reaction is

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First Order Reaction -

The rate of the reaction is proportional to the first power of the concentration of the reaction

- wherein

Formula:

R    \rightarrow        P

a                 0

a-x             x

rate[r]=K[R]^{1}

\frac{-d(a-x)}{dt}=K(a-x)

\frac{-dx}{dt}=K(a-x)  [differentiate rate law]

ln \:[\frac{a}{a-x}]=kt \:(Integrated rate law)

Unit of k=sec^{-1}

t_\frac{1}{2}=\frac{0.693}{k}

 

 

 Let the reaction in first order

then, k=\frac{0.693}{t_{\frac{1}{2}}}= \frac{0.693}{2}= 0.3465 hr^{-1}

Also in first order reactions,

kt = 2.303 log \left ( \frac{a}{a-x} \right )

k = \frac{2.303}{4}  log \left ( \frac{a}{a-\frac{3a}{4}} \right )

k=\frac{2.303}{4}log(4)    = 0.3465 hr^{-1}

Since, the above equation gives the same value of k, so, the reaction is first order.

Alternate method

reaction goes to 50% completion in 2 hrs

 t_{\frac{1}{2}} = 2 hr

further, after 2 hours, reactant goes from \frac{a}{2}  to \frac{a}{4}   in conctraction

Since t_{\frac{1}{2}} =2  hrs in both cases, it means that half life is independent of initial concentration which is the case in first order reaction.

 


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Posted by

Vakul

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