Please show how the exercise above is solved (answer provided). Q: The reaction

Question

Please show how the exercise above is solved (answer provided). Q: The reaction mechanism for the decomposition of A2 A2 -> <- A + A (fast) A + B -> P (slow) involved an intermediate A. Determine the rate law for the reaction in two ways by (a) assuming a pre-equiblibrium and (b) making a steady-state approximation. I don't understand how to arrive at the answer for part (b). Thank you!

21.16(a) The reaction mechanism for the decomposition of A2 A2 A+ A (fast) A +B P (slow) involves an intermediate A. Deduce the rate law for the reaction in two ways by (a) assuming a pre-equilibrium and (b) making a steady-state approximation.

Explanation / Answer

Given mechanism is A2<---> A+A ( fast) and A+B--->P (slow)

in a series of reactions, the slowest will be the rate limiting step.

Hence -dA/dt= K3[A] [B]

but from equilibrium reaction K1[A2] = K2[A]2, [A]= sqrt{(K1/K2)*A2}

rate of reaction -dA/dt= K3*sqrt{( K1/K2)*A2}*B = K3* K1/2K'[B]* sqrt(A2)

under pseudo steady state -dA2/dt= K1[A2]-K2[A]2

but intermediate is A and its net rate is zero hence K1[A2] -K2[A]2 - K3[A][B] =0

K1[A2]-K2[A]2 =K3[A][B]

hence net rate -dA2/dt= K3[A] [B]

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