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### Course: AP®︎/College Chemistry>Unit 7

Lesson 6: Using the reaction quotient

# Worked example: Using the reaction quotient to find equilibrium partial pressures

In some equilibrium problems, we first need to use the reaction quotient to predict the direction a reaction will proceed to reach equilibrium. Once we know this, we can build an ICE table, which we can then use to calculate the concentrations or partial pressures of the reaction species at equilibrium. Created by Jay.

## Want to join the conversation?

• why does partial pressure change in the same ratio as moles? when doing an ICE box with concentrations, the volume stays the same, but isn't it possible for pressure to change throughout the reaction until equilibrium?
(1 vote)
• We can model the behavior of the gases of this reaction using the ideal gas law, PV = nRT. We can rearrange the equation for pressure which yields: P = (nRT)/V. Which means that pressure of the reaction, the total pressure, will change if we also change the temperature, the volume, or the moles of gas. The reaction is at a constant temperature of 1000°C because of the equilibrium constant used. The volume isn’t explicitly described, but if we assume the reaction occurs in a closed rigid container then the volume would also be constant. The total moles of gas also remains the same since the coefficients of gaseous chemicals are both 1. So that means while the partial pressure of carbon monoxide and carbon dioxide may change, the moles disappear and are created respectively at the same rate.

We can also view this behavior using the ideal gas law and Dalton’s law. Dalton’s law saying that the total pressure of a gaseous mixture is the sum of the partial pressures of the components. Using Dalton’s law we can say: Ptot = PCO + PCO2, where Ptot is the total pressure, PCO is the partial pressure of carbon monoxide, and PCO2 is the partial pressure of carbon dioxide. Additionally, we know that pressure is equal to (nRT)/V because of the ideal gas law which means we can rewrite the Dalton law equation as: (ntotRT)/V = ((nCO)RT)/V + ((nCO2)RT)/V, with all the n’s representing the moles of the gases. If we again assume that temperature, volume, and the total moles of gas do not change then we can simplify the previous equation as: constant = nCO + nCO2. So again the sum of the moles of both gases remains constant, but individually they can change (which corresponds to a change in partial pressure). From the chemical equation we know that carbon monoxide and carbon dioxide are related by a 1:1 mole ratio, meaning that the moles of carbon monoxide which are consumed are the same as the moles of carbon dioxide created. All this means that the partial pressures change at the same ratio as the moles of the gases.

Hope that helps.
• I dont see the point of comparing the reaction quotient with the equilibrium pressure, cant you just use an ICE table assuming +x on reactant side and -x on product side, and when you solve for x the signs will balance out to get the equilibrium partial pressures?
(1 vote)
• We need to know the reaction quotient because we need to know whether the production of reactants or products is favored. Essentially we need to know if the reactants/products are increasing or decreasing. Knowing the reaction quotient allows us to know reactant concentrations will increase as opposed to them decreasing. This ultimately gives us the correct information for the ICE table.

Hope that helps.
• If the flow of reaction is from right to left, CO will be the product and CO2 will be the reactant. How come Kp = PCO2/PCO, instead of PCO/PCO2? Thanks.
(1 vote)
• Equilibrium expressions are always defined as the ratio of the products to the reactants. If you take the reciprocal, then you’re writing the equilibrium expression for the reverse reaction and the equilibrium constant changes value.

Using the reaction quotient to find that the reaction is not at equilibrium and needs to shift simply means either the forward or reverse reaction rate increases in order for the reaction to reach equilibrium. It’s the same reaction though so the equilibrium expression remains the same.

Hope that helps.
(1 vote)
• Sorry to ask something completely unrelated to chemistry, but at shouldn't it be -0.192? Because 0.40 - 0.208 = -0.192.