Equilibrium calculations?

Question

Hi!

Can somebody please help me?

At 1200 K, the approximate temperature of vehicle exhaust gases, Kp for the reaction

2CO2(g) -> <- 2CO(g) + O2(g)

is about 1 E-13. Assuming that the exhaust gas (total pressure 1 bar) contains 0.2% CO, 12% CO2 and 3% O2 by volume, is the system at equilibrium with respect to the reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the reaction above?

Thanks a million! :-)

Answer 1

We need to calculate the partial pressures of the three components.

1 bar = 750 mm Hg = 0.986 atm.

At constant T and P, volume and moles are proportional, so volume percent = mole %.

Pa = (xa)(Pt) , , ,the partial pressure of &quot;a&quot; is equal to the mole fraction of &quot;a&quot; times the total pressure. (mole fraction = %)

P CO = (0.002)(0.986) = 0.00197 atm

P O2 = (0.03)(0.986) = 0.0296 atm

P CO2 = (0.12)(0.986) = 0.118 atm

Qp = (P CO)^2 (P O2) / (P CO2)^2 = (0.00197)^2 (0.0296) / (0.118)^2 = 8.3 x 10^-6 which is much greater than Kp = 1 x 10^-13. Since Qp &gt; Kp, the reaction will go back to the left (make more CO2). Adding a catalyst does not change the amount of CO produced, only how fast it is produced.

Answer 2

by assuming that this is ideal gas conditions, the percentage in volume can be related to the percentage in pressure to calculate the partial pressure of each specimen.

P(CO)=0.002 bar

P(CO2)=0.12 bar

P(O2)=0.03 bar

Calculate the reaction quotient, Q

Q=[CO]^2[O2]/[CO2]^2=(0.002)^2*(0.03)/...

Since Q&gt;K, the system is not in equilibrium. Instead, the system will shift to the left. As in, the reaction would occur in the reverse direction forming carbon dioxide.

Rate of reaction does not affect the equilibrium of a reaction. Hence, even if a catalyst is added, the equilibrium of the reaction remains unchanged.