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the gaseous reaction A_{2}---> 2a is first order in A_{2} .after 12.3 minutes 65% of A_{2} remains undecomposed . how long will it take to decompose 90% of A_{2} ?what is the half life of the reaction?

As this is a 1st order reaction,

So, ln A = -kt + ln A

_{o }

i.e. ln (0.65) = - (k) (738 s) + ln 1.00 [65% means 0.65 if total conc. initially was 1 i.e. A

_{o}= 1, 12.3 min = 738 s]

or, k = 0.000584010 s

^{-1}.

Now, to decompose 90% means 10% remains. So, putting this in equation,

ln A = -kt + ln A

_{o}

ln 0.100 = - ( 0.000584010 s

^{-1}) (t) + ln 1.00

t = 3943 s i.e. almost 66 minutes.

Half life of reaction = $\frac{0.693}{k}=\frac{0.693}{0.000584010}=1187sie.almost20minutes.$

Regards

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