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A first order reaction is `40%` completed in 50 minutes. What is the time required for `90%` of the reaction to complete? |
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Answer» For the first order reaction, `t=(2.303)/t log a/(a-x)` Ist case: `a=100%, x=40%, (a-x)=100-40=60%, t=50` min `t_(40%)=(2.303)/k log(100%)/(60%)` `50"min" = 2.303/k log (100)/(60)=2.303/k log 1.667 = (2.303 xx 0.2219)/k`…………(i) IInd case: `a=100%, x=90%, (a-x)=100-90=10%` `t_(90%) = 2.03/k log (100%)/(10%)=2.303/k log 10=2.303/k`.........(ii) Divide eqn. (ii) by eqn. (i), `t_(90%)/(50"min") = 2.303/(2.303 xx 0.2219) = 4.506 t_(90%)=4.506 xx (50"min") = 2.25 xx 10^(2)`min. |
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