11/10 in-class activity

For the reaction of solid carbon (C_(s)) and gaseous carbon dioxide (with the product gaseous carbon monoxide), K_p=1.50.

1.) Write the balanced chemical equation (remember this is an equilibrium).

                                          k₁

            C_(s) + CO_2,(g)    ßà       2 CO_(g)

                                     k_-1

2.) Write the forward and reverse rate laws in terms of partial pressure.  Come up with an expression for equilibrium.

Forward rate = k₁ (Pco₂)

Reverse rate = k_-1 (Pco)²

k₁ is the forward reaction rate constant and k_-1 is the reverse reaction rate constant.

Equilibrium constant, K_p = (Pco)²/(Pco₂)

Remember that solids and pure liquids are not included in ICE diagrams of equilibrium expressions (it doesn’t make sense to have a concentration or a partial pressure of a solid or pure liquid).

3.) For excess C_(s) and initial partial pressures Pco₂ = 0.56 atm and Pco = 0.14 atm, find Q_p, the reaction quotient.  Which way will the reaction proceed to equilibrium?

Q_p = (Pco)²/(Pco₂) = 0.035

Q<K so the reaction will proceed to the right (products) to reach equilibrium.

4.) Make an ICE diagram for this reaction using the partial pressures.

	CO2            à	2CO
I (initial)	0.56 atm	0.14 atm
C (change)	-x	+2x
E (equilibrium)	0.56 - x	0.14 + 2x

5.) Solve for x.  You will get two values for x here.  Which one makes physical sense?

K_p = 1.50 (given in the problem)

Given the equilibrium constant expression K_p = (Pco)²/(Pco₂), we can use the new equilibrium concentration expressions (found in the ICE table under equilibrium).  Plugging this in, we get:

1.50 = (0.14 + 2x)²/(0.56 – x)

multiplying both sides by (0.56 – x):

1.5*(0.56 – x) = (0.14 + 2x)²

0.84 – 1.50x = 0.0196 + 0.56x + 4x²

Therefore, 0 = 4x² + 2.06x – 0.8204

Plugging into the quadratic equation (see video), we get:

x = 0.263 atm and x = -0.778 atm

We know that the partial pressure of CO₂ will decrease, so x must be positive.  Also, for x = -0.778, we get an equilibrium pressure of CO of 0.14 + 2*-0.778, which doesn’t make physical sense (we can’t have a negative partial pressure).  Therefore, we get x = 0.263 atm.

6.) If you add 0.042 mol CO to the 2L system at 293 K at equilibrium, what is Q_p?  According to Le Châtelier’s principle, to which direction will the reaction proceed?

For x = 0.263 atm, we get equilibrium partial pressures of: Pco = 0.666 atm and Pco₂ = 0.297 atm.  Putting the added number of moles into the equation for the ideal gas law (PV=nRT), we get a partial pressure of 0.505 atm.  We add this to the partial pressure of CO we already have and plug it into the expression for Q_p:

Q_p = (0.505 + 0.666)²/(0.297) = 4.62

Q>K, so the reaction will proceed to the left (toward reactants).  Le Châtelier’s principle also tells us that when we add more products to a system at equilibrium, the system will shift toward reactants.