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Following is the graph between `(a-x)^(-1)` and time for second order reaction `Ato2B,Q=tan^(-1)(0.5),OA=2Lmol^(-1)`. Here is the initial concentration of reactant and x is its amount reactedc at any time t. Hence rate at the start of the reaction is:A. `0.125Lmol^(-1)mol^(-1)`B. `0.25molL^(-1)min^(-1)`C. `0.125molL^(-1)min^(-1)`D. `0.25Lmol^(-1)min^(-1)` |
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Answer» Correct Answer - C For second order reaction, `because(a-x)^(-1)=(1)/(Kt)+(I)/(a)` `becausek=tantheta(becausetantheta=0.5` from question) `becauseK=0.5` lit `mol^(-1)min^(-1)` and `(I)/(a)=2` or `a=0.5M` `r=Ka^(2)` `=0.5xx0.5^(2)` `=0.125molL^(-1)min^(-1)` |
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