How much energy in Joules) would it take to warm 3.11 grams of gold by 7.3°C (specific heat capacity of gold = 0.130 J/g °C)

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]