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The rate of a first order reaction is `0.04 mol l^(-1) s^(-1)` at 10 seconds and `0.03` mol `l^(-1) s^(-1)` at 20 seconds after initiation of the reaction . The half-life period of the reaction isA. 24.1 sB. 34.1 sC. 44.1 sD. 54.1 s |
Answer» Correct Answer - a `K = (2.303)/((t_(2) - t_(1))) "log" ((a-x_(1)))/((a-x_(2))) K = (2.303)/((20-10)) "log" ((0.04)/(0.03))` `K = (2.303 xx 0.1249 )/(10) (2.303 xx "log" 2)/(t_(1//2) = (2.303 xx "log" 2)/(10)` `t_(1//2) = (0.3010 xx 10)/(0.1249) = 24.1` sec |
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