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Question

A first order gaseous reactions has K=1.5×106sec1 at 200C. If the reaction is allowed to run for 10 hour, what percentage of initial concentration would have changed into products. What is the half life period of reaction?

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Solution

K=2.303tlogAoA
1.5×106×10×60×602.303=log(AoA)
Ao=1.05A
Ao=1.05(Aox)
1.05x=1.05AoAo
x=0.051.05Ao
x=0.0477Ao
x=4.7% Ao
t1/2=0.6931.5×106
t1/2=4.6×105sec

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