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.

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]

the objects have the same charge is correct.

nathan lee