Question

Suppose the formation of nitrosyl chloride proceeds by the following mechanism: step elementary reaction rate constant 1 NO(g) + Cl2(g) → NOCl2(g) k1 2 NOCl2(g) + NO(g) → 2NOCl(g) k2 Suppose also k1≫k2. That is, the first step is much faster than the second. Write the balanced chemical equation for the overall chemical reaction: Write the experimentally-observable rate law for the overall chemical reaction. =ratek Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of k1, k2, and (if necessary) the rate constants k-1 and k-2 for the reverse of the two elementary reactions in the mechanism. =k

Answer 1

Answer:

Overall reaction equation;

2NO(g) +Cl2(g) -----> 2NOCl (g)

Explanation:

Given

1) NO(g) + Cl2(g) → NOCl2(g)

2) NOCl2(g) + NO(g) → 2NOCl(g)

Overall reaction equation;

2NO(g) +Cl2(g) -----> 2NOCl (g)

k1= [NOCl2]

k-1= [NO] [Cl2]

k2 = [NOCl2] [NO]

Equilibrium for the first equation (reaction 1)

K= k1/k-1 = [NOCl2]/[NO] [Cl2]

Therefore

[NOCl2] = k1/k-1 [NO] [Cl2]

Rate= k2× k1/k-1 [NO]^2 [Cl2]

Rate = Koverall [NO]^2 [Cl2]

Where Koverall= k1k2/k-1