How does decreasing temperature effect equilibrium

In considering volume changes, we must pay close attention to the phases that are affected by the change. If there are aqueous, solid, and gas phases in a container, changing the volume of the container only changes the volume of the gas phase.

Adding water to the container obviously increases the volume of the aqueous phase, but the effect on the gas phase depends on whether the addition is at constant volume or constant pressure. If the addition is at constant pressure the volume of the gas phase is not changed , but if the volume of the container is kept constant the volume of the gas phase decreases as the volume of the aqueous phase increases.

Addition of an immiscible liquid such as mercury has no effect at constant pressure , but decreases the volume of the gas phase if the volume of the container is constant. Addition of an inert gas has no effect if the volume of the container is constant, but increases the volume of the gas phase if the pressure is constant.

Changing the volume of a phase changes the concentrations of all components in that phase, reactants and products. The net effect is determined by the difference in the number of molecules of reactants and products from the balanced reaction in that phase.

If the balanced reaction shows the same number of moles of reactants and products in the phase being changed, there is no effect on the equilibrium. If the balanced reaction shows more moles of products than reactants in the phase being changed, the effect on the products predominates.

Therefore, if the volume of that phase is increased the concentrations of products will be decreased more than the reactants, and the equilibrium will shift to increase the amounts of products and decrease the amounts of reactants.

A decrease in the volume of that phase would have the opposite effect. This is typical of what happens with any equilibrium where the forward reaction is exothermic.

Increasing the temperature decreases the value of the equilibrium constant. Where the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant. The position of equilibrium also changes if you change the temperature.

If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favoring the reaction which absorbs heat. So, according to Le Chatelier's Principle the position of equilibrium will move to the left with increasing temperature.

Less hydrogen iodide will be formed, and the equilibrium mixture will contain more unreacted hydrogen and iodine. That is entirely consistent with a fall in the value of the equilibrium constant. Equilibrium constants are not changed if you add or change a catalyst.

The position of equilibrium is not changed if you add or change a catalyst. A catalyst speeds up both the forward and back reactions by exactly the same amount. Dynamic equilibrium is established when the rates of the forward and back reactions become equal. That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible.

It can do this by favouring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium. Let's look at the same equilibrium we've used before.

This one would be affected by pressure because there are 3 molecules on the left but only 2 on the right. An increase in pressure would move the position of equilibrium to the right. Once again, it is easy to suppose that, because the position of equilibrium will move to the right if you increase the pressure, K p will increase as well.

Not so! Note: If you aren't happy with this, read the beginning of the page about K p before you go on. If you sort this out, most of the "P"s cancel out - but one is left at the bottom of the expression. Now, remember that K p has got to stay constant because the temperature is unchanged. How can that happen if you increase P? To compensate, you would have to increase the terms on the top, x C and x D , and decrease the terms on the bottom, x A and x B. Increasing the terms on the top means that you have increased the mole fractions of the molecules on the right-hand side.

Decreasing the terms on the bottom means that you have decreased the mole fractions of the molecules on the left. That is another way of saying that the position of equilibrium has moved to the right - exactly what Le Chatelier's Principle predicts.

The position of equilibrium moves so that the value of K p is kept constant. There isn't a single "P" left in the expression. Changing the pressure can't make any difference to the K p expression. The position of equilibrium doesn't need to move to keep K p constant. Equilibrium constants are changed if you change the temperature of the system. K c or K p are constant at constant temperature, but they vary as the temperature changes.

When new bonds are generated, more thermal energy is released that needed to break bonds in the reactants. In this chemical reaction. Raising the temperature favors the reverse reaction endothermic and similarly Lowering the temperature favors the forward reaction exothermic. Le Chatelier's principle explains that the reaction will proceed in such a way as to counteract the temperature change.

The exothermic reaction will favor the reverse reaction, opposite the side heat is the opposite is true in endothermic reactions; the reaction will proceed in the forward reaction.

Although it is not technically correct to do so, if heat is treated as product in the above reaction, then it becomes clear that if the temperature is increased the equilibrium will shift to the left using Le Chatelier's principle.

If temperature is decreased, the reaction will proceed forward to produce more heat which is lacking.