InterviewSolution
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InterviewSolution
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A reaction is of second order with respect to its reactant. How will its reaction rate be affected if the concentration of the ractant is (i) doubled (ii) reduced to half ? |
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Answer» Since Rate = K `[A]^(2)` Let [A] = a `" " therefore " " Rate = Ka^(2)…. (1)` (i) `" "` If [A] = 2a `" " therefore " "` Rate = K `(2a^(2)) = 4 Ka^(2) = 4 ` times (ii) `" "` If [A] = `(a)/(2) " " therefore " " Rate = K ((a)/(2))^(2) = (1)/(4)` `Ka^(2) = (1)/(4)` th |
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