s0. Extra Credit Problem (6 points possible) In a reversible reaction that reach

Question

s0. Extra Credit Problem (6 points possible) In a reversible reaction that reaches equilibrium, the equilibrium constant, becomes a constant value. Suppose the following reaction occurs as a concerted one-step process, with no intermediates, both in the forward and reverse steps. so, (g) + SO (g) 2 SO2 (g) k.i Use algebra, the rate laws for the forward and reverse reactions, and the definition of the equilibrium constant K, to show how the equilibrium constant is a true constant, and how it is mathematically related to the microscopic rate constants for the forward and reverse reactions (k1 and k.1).

Explanation / Answer

Kc = [SO2[^2/[SO3][SO]

with k1 and k-1 be rates for forward and reverse reaction

-d[SO3]/dt = k1[SO3][SO] = -k-1[SO2]^2

[SO2]^2 = k1/k-1[[SO3][SO]

Feed in Kc

Kc = (-k1/k-1)[SO3][SO]/[SO3][SO]

     = -k1/k-1 = k-1/k1 = constant

So, we have shown that the equlibrium constant Kc is a constant.

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