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Experiments show that each of the following redox reactions is second-order overall: (1) NO2(g) + CO(g) → NO(g) + CO2(g) (2) NO(g) + O3(g) → NO2(g) + O2(g) (a) When [NO2] in reaction 1 is doubled, the rate quadruples. Determine the rate law for this reaction. Rate = k[NO2][CO] k[NO2]2 k[NO2]3[CO]−1 k[NO2]4[CO]−2 (b) When [NO] in reaction 2 is doubled, the rate doubles. Determine the rate law for this reaction.

Respuesta :

Answer:

a) rate law1 = k[NO2]²

b) rate law2 = k[NO][O3]

Explanation:

NO2(g) + CO(g) → NO(g) + CO2(g)

NO(g) + O3(g) → NO2(g) + O2(g)

When [NO2] in reaction 1 is doubled, the reaction quadruples

Rxn is second order.

rate law1= [NO2]^a [CO]^b

rate law1= [NO2]² [CO]^0

rate law1 = k[NO2]²

When [NO] in reaction 2 is doubled, the rate doubles.

Rxn is first order

The ratio is 1:1

this makes the rate law2 = k[NO][O3]

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