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InterviewSolution
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Following is the graph between `(a-x)` and time `t` for second order reaction `theta=tan^(-1)(0.5)OA=2L mol^(-)` Hence, the rate at the start of the reaction isA. `1.25 mol L^(-1) mi n^(-1)`B. `0.5 mol L^(-1)mi n^(-1)`C. `0.125 mol L^(-1) mi ^(-1)`D. `12.5 mol L^(-1) min^(-1)` |
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Answer» Correct Answer - C `kt=((1)/(a-x)-(1)/(a))` `(1)/(a-x)=kt+(1)/(a)` Graph between `(a-x)^(-1)` and time `t` is a straight line, hence, `k= tan theta= 0.5` `OA=(1)/(a)=2` `:. A=(1)/(2)=0.5` `:. R=k[A]^(2)` `= 0.5xx(0.5)^(2)` `=0.125Lmol^(-1)min^(-1)` |
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