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Question

# (a) A reaction is second order in A and first order in B.(i) Write the differential rate equation.(ii) How is the rate affected on increasing the concentration of A three times?(iii) How is the rate affected when the concentrations of both A and B are doubled?(b) A first order reaction 40 minutes for 30% decomposition. Calculate t1/2 for this reaction. (Give log 1.428 = 0.1548)

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Solution

## Given that Order of reactant A, x=2Order of reactant B,y=1The differential rate equation will be−d[R]dt=k[A]x[B]yPlug the value in this formula−d[R]dt=k[A]2[B]Given that concentration of A is increased three timesSo that New concentration of A =3[A]Put the values we get new rate of reaction−d[R]dt=k[3A]2[B]Hence, the rate of reaction will increase by 9 times.(iii)Given that New concentrations of A=2[A]New concentration of B=2[B]Plug the values we get−d[R]dt=k[2A]2[2B]Hence the rate of reaction increased by 8 times (b) k=1tln100100−30 =140×0.1548 =0.00387min−1

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