A reaction is of first order in a reactant A and of second order in a reactant B . How is the rate of this reaction affected when
a) The concentration of B alone is increased to three times?
b)The concentration of A as well as B are doubled ?
1
Expert's answer
2012-10-12T08:21:55-0400
Since it is given that a reaction is first order in reactant A and second order in reactant B; therefore, r = k[A][B]2 …………………………. (i)
Where r is the rate of reaction and k is the rate constant of the reaction.
a) When concentration of B alone is increased three times, let the new rate be r1 r1 = k[A][3B]2 = 9k[A][B]2 …………………………. (ii) Dividing eq.(ii) by eq.(i), we get r1 = 9r Therefore, the rate of the reaction would become 9 times the initial rate when the concentration of B alone is increased three times.
b) When the concentration of both the reactants is doubled, then the rate of the reaction would be r2 r2 = k[2A][2B]2 = 8k[A][B]2 …………………………. (iii)
Dividing eq.(iii) by eq.(i), we get r2 = 8r
Therefore, the rate of the reaction would become 8 times the initial rate when the concentration of both A and B is doubled.
Comments
Leave a comment