Question

Consider the reaction: 2 NH3 (aq) + OCl-(aq) -> N2H4(aq) + H2)(I) + Cl-(aq) This three-step...

Consider the reaction: 2 NH3 (aq) + OCl-(aq) -> N2H4(aq) + H2)(I) + Cl-(aq)

This three-step mechanism is proposed:

Nh3(aq) + OCl-(aq) --><-- NH2Cl(aq) + Cl-(aq) Fast

NH2Cl(aq) + NH3(aq) --> N2H5+(aq) + Cl-(aq) Slow

N2H5+ (aq) + OH-(aq) --> N2H4 (aq) + H2O(l) Fast

A. Show that the mechanism sums to the overall reaction.

B. What is the rate law predicted by the mechanism?

The answer is wrong in the textbook solutions, so I can't check my work and it's driving me nuts!! No online answers here show the steps. :(

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Homework Answers

Answer #1

Part b

Rate is determined by the slowest step

2nd reaction is slow

Rate law for second reaction = k [NH2Cl][ NH3]

But NH2Cl is intermediate

Rate law for first reaction = = k [OCl-][ NH3]

rate law for reaction 1 into rate law for reaction 2

Replacing NH2Cl intermediate

Rate = k [ NH3][OCl-][ NH3]

Rate = k[NH3]^2[OCl-]

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