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A reaction is of second order with respect to a reactant . How is its effected if the concentration of the reactant is (i)doubled (ii) reduced to half ?A. Becomes 2 times and `(1)/(2)` times respectivelyB. Becomes` (1)/(2) "timses and" (1)/(2)` times respectivelyC. Becomes 2 times and 4 times, respectivelyD. Becomes 4 time and 2 times, respectively |
Answer» (For first order reaction ), rate =k [A] we have to find the rate expression when conecentration is doubled or reduced to half and compare it with the normal rate. E.g. for a reaction A `to` products Rate =k[A]=ka (i) When the concentration of A is doubled [A]=2a Rate =k (2a)=k2a Rate of reaction becomes 2 times (ii) When concentration of Ais reduced to `(1)/(2)` ` [A] =(1)/(2)a` Rate = `k ((a)/(2)) =(ka)/(2)` Rate of reaction becomes `(1)/(2)` times i.e reduced to twice |
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