InterviewSolution
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InterviewSolution
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A second order reaction requires `70 min` to change the concentration of reactants form `0.08 M` to `0.01 M`. The time required to become `0.04 M = 2x min`. Find the value of `x`. |
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Answer» Correct Answer - 5 For second order reaction: `[R]_("initial") = 0.08, M, [R]_("initial") = 0.01M` `x = 0.08 - 0.01 = 0.07M` `:. (a-x) = 0.08 - 0.07 = 0.01 M` `k_(2) = (1)/(t).(x)/(a(a-x))` `= (1)/(70 min) xx (0.07 M)/(0.08M xx 0.01 M)` …(i) Now, time required to become concentration `= 0.04 M`. i.e., `x = 0.04 M` `k_(2) = (1)/(t) xx (0.04 M)/(0.08 M xx (0.08-0.04)M)` ...(ii) form Eqs. (i) and (ii) `(0.07)/(70 xx 0.08 xx 0.01) = (0.04)/(t xx 0.08 xx 0.04)` `t = 10 min = 2x min` `:. x = 5 min`Correct Answer - 5 For second order reaction: `[R]_("initial") = 0.08, M, [R]_("initial") = 0.01M` `x = 0.08 - 0.01 = 0.07M` `:. (a-x) = 0.08 - 0.07 = 0.01 M` `k_(2) = (1)/(t).(x)/(a(a-x))` `= (1)/(70 min) xx (0.07 M)/(0.08M xx 0.01 M)` …(i) Now, time required to become concentration `= 0.04 M`. i.e., `x = 0.04 M` `k_(2) = (1)/(t) xx (0.04 M)/(0.08 M xx (0.08-0.04)M)` ...(ii) form Eqs. (i) and (ii) `(0.07)/(70 xx 0.08 xx 0.01) = (0.04)/(t xx 0.08 xx 0.04)` `t = 10 min = 2x min` `:. x = 5 min` |
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