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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)

(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=2

Order of reactant B,y=1

The differential rate equation will be

Order of reactant A, x=2

Order of reactant B,y=1

The differential rate equation will be

−d[R]dt=k[A]x[B]y

Plug the value in this formula

−d[R]dt=k[A]2[B]

Given that concentration of A is increased three times

So 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

Plug the value in this formula

−d[R]dt=k[A]2[B]

Given that concentration of A is increased three times

So 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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