Adding an inert gas to the system does not change the equilibrium as it does not shift the direction of equilibrium either direction. It doesn't shift the equilibrium because the partial pressures of the gases remain the same. The case for random gases depends upon what you mean by random. If the "random" gas is HF, it has an effect. If CF4 or CO2, either will have an effect. But a gas that doesn't enter into the reaction , such as argon, or helium, or any gas other than CO2, HF, or CF4, there will be no effect because the partial pressures of HF, CO2, and CF4 are not changed..
For the case of water, if H2O was a gas and entered into the reaction, you could make a case for removing water and shifting the equilibrium to the left. But since it is stated in the equation that H2O is a liquid, that is its normal state, it has an activity of 1, AND AS LONG AS THERE IS A DROP OF LIQUID WATER PRESENT, adding or removing liquid water does not change the concentration of liquid water and there is no effect on the equilibrium.
Given the equilibrium rxn
CF4(g) + 2H2O(l) ⇔ CO2(g) + 4HF(g) ΔH < 0
if pressure is increased by adding argon gas to the equilibrium, I know it doesn't decrease the amount of HF. But if a random gas is added for any equilibrium rxn, will it always cause an increase in the products? Do we always assume it's being added to reactants?
Also if Water is removed, why doesn't it decrease the amount of HF? I thought it would shift to the left if water was removed, causing the products to decrease?
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