Consider the following proposed mechanism for the reaction A rightarrow P Write
ID: 1022615 • Letter: C
Question
Consider the following proposed mechanism for the reaction A rightarrow P Write a differential equation for the rate of formation of B. b) The formation of B from A and C is fast to equilibrium with equilibrium constants K_1 = k_1/k_-1, K_2 = k_2/k_-2, and K_3 = k_3/k_-3. Also, k_4 is smaller that the other k's. Write a differential equation for the formation of P in terms of concentration of A, equilibrium constants, and k_4. c) Write an expression, containing no derivatives, for the concentration of P as a function of time. At zero time, [A] = [A_0] and [B], [C], [P] = 0.Explanation / Answer
In the reaction, B is formed through decompositino of A and C and decomposed to A , C and P.
dB/dt= K1[A]- K-1[ B] + K2[C]-K-2[B]
since formation of B from A and C is fast
k1[A] = k-1[B] . [B]/[A] = k1/k-1= K1, [B] = K1[A]
k3[A] = k-3[C], [C]/[A] = k3/k-3= K3, [C] = K3[A]
k2[C]= k-2[B], [B]/[C] = k2/k-2= K2, [B] = K2[C], [B]= K2K3[A]
Since the slowest step determines the rate
dP/dt= K4[B] =k4[ K1[A] + K2K3[A] = k4[A] [K1+K2K3]
Ao= A+B+C+P ( C , P, A and B are concentrations at any time t)
P= A0-B-C