Le Chatelier's Principle
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- เผยแพร่เมื่อ 22 ส.ค. 2024
- Need help preparing for the General Chemistry section of the MCAT? MedSchoolCoach expert, Ken Tao, will teach you Le Chatelier's Principle for kinetics. Watch this video to get all the mcat study tips you need to do well on this section of the exam!
Recall that a reaction is at chemical equilibrium when the rate of the forward reaction (product formation) is equal to the rate of the reverse reaction (reactant formation). Le Chatelier's principle states that for a reaction at equilibrium, a stress applied to that reaction will be compensated for by the reaction in order to minimize that stress. Ways of stressing a reaction include changing the ratio of reactants to products, changing the volume or pressure of the reaction environment, or heating the reaction system. Stressing a system forces it into an energetically unfavorable state. The chemical reaction may shift toward reactant formation or product formation in order to “relieve” that stress. A list of the most common stresses and corresponding reaction shift is shown below.
Applications of Le Chatlier’s
Consider the equilibrium state of the following reaction, noting once again that solids are not included in our expression for equilibrium. While Keq is the ratio of products to reactants at equilibrium, we define another value Q to be the ratio of products to reactants at any given moment in time in our reaction system. If we know the value of Q does not equal Keq, the reaction will proceed in whatever direction allows it to return to equilibrium.
Here are some specific instances of Le Chatlier’s law at work, using the above reaction:
1. If the reaction is at equilibrium, adding excess [CO] increases the ratio of products to reactants in the system. Therefore, Q greater than Keq. The reaction will want to reduce the number of products and increase the number of reactants, so it will favor the reverse reaction.
2. If two moles of solid carbon are added, no effect is had on reaction equilibrium, because that substance is a solid and therefore a nonparticipant in determining reaction equilibrium.
3. If the volume of the container of the reaction system is doubled, pressure will become half of its original value. The reaction will seek to compensate for this by increasing pressure. Recalling the ideal gas law, we know that to increase pressure, the number of moles of gas in the system must be increased. To do so, the reaction will shift to the right, because 2 moles of gas (H2 and CO) are present on the product-side, whereas only 1 mole of gas is present on the reactant-side.
4. If the temperature of the reaction is decreased, the energy of the system can be said to have decreased. The reaction will seek to “release” energy to compensate for this. It can accomplish this by favoring the exothermic reaction, or the one that releases heat energy. In this case, the forward reaction is exothermic, producing heat. Therefore, the forward reaction is the one that will be favored when temperature decreases.
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amazing explanation! thank you
for part E could you tell me how you found that the reaction is exothermic?
This is a little late but basically he said that you needed more information to answer part E. He choose exothermic arbitrarily. On the MCAT they would tell you before hand whether the reaction is exothermic or endothermic or you would be given enough info to calculate it yourself.