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For the first order reaction, `k=5.48 xx 10^(-4)s^(-1)`, the two-third life time for this reaction isA. 2005 sB. 1000 sC. 2000 sD. 3005 s |
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Answer» Correct Answer - A For first order reaction, we know that `k=(2.303)/(t).log.(a)/((a-x))` Hence, `t=(2.303)/(5.48 xx 10^(-4))xxlog .(a)/(a-(2)/(3)a)` `=(2.303xx10^(4))/(5.48)xxlog3` `=(2.303xx10^(4))/(5.48)xx0.4771=2005s` |
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