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Question

#1. Consider the following equilibrium: C(s) + CO2(g) 2 CO(g) Kp = 167.5 at 1000°C.
a) (8 marks) Suppose that a flask is charged with some solid carbon plus 0.25 atm of CO2 and 0.25 atm CO.
What will be the partial pressure of each gas when the sample reaches equilibrium?

b) (8 marks) Briefly explain what would happen, and why (not just Le Châtelier’s rule) in this equilibrium
mixture if…
 (i) some CO2 was removed?
(ii) the reaction vessel’s volume was suddenly increased?
(iii) the temperature was increased?
(iv) some (but not all) of the C was removed?

hcbiochem · Accepted Answer

Kp = (PCO)^2 / PCO2 = 167.5
Under the original conditions,
Q = (0.25)^2 / (0.25) = 4
Because Q < K, the reaction will proceed to the right consuming CO2 and forming additional CO. So, let 2x = pressure increase in CO and -x = pressure decrease in CO2. Then,

Kp = 167.5 = (0.25+2x)^2 / (0.25-x)
41.875 - 167.5x = 0.0625 + 1x + 4x^2
4x^2 +168.5x - 41.8125 = 0

Using the quadratic formula gives the positive root of x = 0.2467

So, at equilibrium,
PCO = 0.25 + 2x = 0.742 atm
PCO2 = 0.25 - 0.2467 = 3.3X10^-3 atm

b) (i) The rate of the forward reaction is decreased compared to the rate of the reverse reaction. So, the PCO2 increases until equilibrium is re-established.

(ii) Increasing the volume decreases the pressure of both gases. Because P(CO) is squared in the expression of Kp, the reaction quotient is decreased. Consequently the forward reaction will occur at a higher rate than the reverse reaction. PCO2 will decrease and PCO will increase.

(iii) Without knowing whether this reaction is endothermic or exothermic, no prediction can be made.

(iv) Removing only some of the C(s), the equilibrium is not affected at all.

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