9. Data for the following reaction are given in the table below. CO(g) + NO2(g)
ID: 701750 • Letter: 9
Question
9. Data for the following reaction are given in the table below. CO(g) + NO2(g) CO2(g) + NO(g) Experiment [CO] (mol/L) NO2J(mol/L) Initial rate (Mhr) 5.0 x 10 5.0 x 10 1.0 x 10 1.5 x 10 3.6 x 10 1.8 x 105 3.6 x 10 7.2 x 10 3.4x 10 1.7x10 6.8 x 10 a) What is the rate law for this reaction? b) What is the initial rate for the reaction in experiment 42 10. Consider the following equilibrium at -78.15 °c. The partial pressure of each gas is 0.278 atm. Calculate the equilibrium constant, K for the reaction 195KExplanation / Answer
Ans 9
Let
Rate law
r = k [CO]a [NO2]b
From exp 1 and 2
r1/r2 = [CO]1a [NO2]1b / [CO]2a [NO2]2b
3.4/1.7 = (3.6/1.8)b
2 = 2b
b = 1
From exp 1 and 3
r1/r2 = [CO]1a [NO2]1b / [CO]2a [NO2]2b
3.4/6.8 = (5*10^-4/1*10^-3)a
0.5 = 0.5a
a = 1
Rate law for this reaction
r = k [CO] [NO2]
From exp 1
k = r1/[CO]1 [NO2]1
= (3.4*10^-8)/(5*10^-4*3.6*10^-5)
= 1.8888 M-1 hr-1
Part b
Rate of reaction in experiment 4
r4 = k [CO]4 [NO2]4
= 1.8888 x (1.5*10^-3) x (7.2*10^-5)
= 2.04 x 10^-7 M/hr
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