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A first order reaction is carried out...

A first order reaction is carried out starting with 10 mol `l^(-1)` of the reactant. It is 40% complete in 1h if the same reaction is carried out with an initial concentration of 5 mol `l^(-1)` the percentage of the reaction that is completed in 1h will be

A

`40%`

B

`80%`

C

`20%`

D

`60%`

Text Solution

Verified by Experts

For first order reaction, `t = (2.303)/(k) log((a_(0))/(a_(0)-x))`
Case I : `a_(0) = 10 mol L^(-1)`, For 40% completion of the reaction, `t_(1) = (2.303)/(k) log((10)/(6)) = 1h`
Case II : `a_(0) = 5 mol L^(-1)`, If the initial concentratin, that is , `a = 5 mol L^(-1)`, then `t_(2) = (2.303)/(k)log((5)/(5-x))`
Given that `t_(1) = t_(2) = 1h`. So, `log((5)/(3)) = log((5)/(5-x)) ` or `(5)/(3) = (5)/(5-x)` or x = 2
Therefore, the percentage of reaction completed is `(2)/(5) xx 100 = 40%`
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