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When a mixture is compressed, causing the volume to decrease, and there are more moles of gas on the reactant side of the equilibrium equation, which way does the equation shift and why?
Le Chatliers principle states that a system wants to reduce the stress that is applied. As the volume was decreased so the partial pressures were increased, the equilibrium wants to shift to the side with less moles of gas to reduce the pressure. This means it forms more product and favors the right.
I think that increasing the partial pressure of a gas that doesn't participate in a reaction such as He has a similar effect as increasing the pressure of the system, assuming the volume of the reaction vessel is kept constant. If the partial pressure of He is increased without any of the other partial pressures changing, the system's pressure will increase, and it will act to counter that change by shifting the equilibrium towards the side of the reaction with fewer moles of gas.
When you add an inert gas like Helium, (which is NOT needed for the reaction), there will be no effect on the equilibrium constant because there is no change in concentration.
When a mixture is compressed, there is less volume for gasses to exist so the system will shift towards the side with fewer moles of gas. In other words, gasses take up less space so when there is less space, the reaction will go more towards whatever side has fewer moles of gas.
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