For the reaction A ↔ B, we can readily write the differential equations for A and B:
dA/dt = - kfA + krB
dB/dt = + kfA - krB
dB/dt = + kfA - krB
Assuming that there is only A initially, we can state that Ao = A+B, or B = Ao - A, which, when plugged into equation 1 above, makes it integrable (two variables only). The solutions are:
A = Ao[ (kr + kfe-(kf+kr)t)/(kr + kf) ]
B = Ao[ 1 - (kr + kfe-(kf+kr)t)/(kr + kf) ]
B = Ao[ 1 - (kr + kfe-(kf+kr)t)/(kr + kf) ]
To connect to equilibrium, we recognize that A(∞)=Aeq and that B(∞)=Beq:
Aeq = Ao[ kr/(kr + kf) ]
B = Ao[ kf/(kr + kf) ]
B = Ao[ kf/(kr + kf) ]
Cool, from just kinetics we can predict equilibrium quantities of A and B. But not only that, we can take their ratio and obtain the equilibrium constant:
K = kf/kr
Remarkably, this is the result for all reversible reactions ( 2A ↔ B, 3A ↔ 2B, etc) and is called the principle of microscopic reversibility.
More complicated mechanisms can be understood as combinations of branching, sequential and opposing reactions. When intermediates are reactive, we can often use the steady-state approximation, which says that dI/dt ≅ 0. This allows us to more easily obtain rate laws from mechanisms without solving elaborate series of differential equations that take lots of time and make your soul cry.
More complicated mechanisms can be understood as combinations of branching, sequential and opposing reactions. When intermediates are reactive, we can often use the steady-state approximation, which says that dI/dt ≅ 0. This allows us to more easily obtain rate laws from mechanisms without solving elaborate series of differential equations that take lots of time and make your soul cry.
2 comments:
FYI your debye-huckel theory and sedimentation handout links, for the course page, do not work.
Hey Dr. Schoonover,
Could you post hw#6 on the website... I've somehow misplaced mine!!
Thanks,
Amanda
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