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For the equilibrium system below, which of the following would result in an increase in the concentration of CO(g)?

2H2(g) + CO(g) CH3OH(g) + 92 kJ
a) decreasing temperature
b) adding some CH3OH
c) both b and d
d) decreasing the volume of the container
e) both a and b

To determine which option would result in an increase in the concentration of CO(g) in the given equilibrium system, let's analyze each option:

a) Decreasing temperature: According to Le Chatelier's Principle, when the temperature is decreased in an exothermic reaction, the reaction tends to shift in the forward direction to compensate for the decrease in temperature. Since the reaction given is exothermic (indicated by the release of 92 kJ), decreasing the temperature would shift the equilibrium towards the reactants. Thus, decreasing the temperature would result in an increase in the concentration of CO(g).

b) Adding some CH3OH: Adding more CH3OH to the system would shift the equilibrium towards the products, as the increase in the concentration of one of the products would cause the system to try to reach equilibrium again. This means that adding CH3OH would result in a decrease in the concentration of CO(g).

c) Both b and d: Option c suggests that both adding CH3OH and decreasing the volume of the container (which has the same effect as increasing the pressure) would result in an increase in the concentration of CO(g).

d) Decreasing the volume of the container: According to Le Chatelier's Principle, when the volume of a system is decreased, it increases the pressure. In this equilibrium system, decreasing the volume would increase the pressure and according to Le Chatelier's Principle, the reaction would shift in the direction that produces fewer moles of gas. In the given reaction, there are fewer moles of gas on the product side (CH3OH). Therefore, decreasing the volume would shift the equilibrium towards the products and result in a decrease in the concentration of CO(g).

e) Both a and b: Option e specifies that both decreasing the temperature and adding some CH3OH would result in an increase in the concentration of CO(g). However, as discussed above, adding CH3OH would actually decrease the concentration of CO(g). Therefore, option e is incorrect.

In summary, the correct answer is option a) Decreasing temperature.

To determine which of the given options would result in an increase in the concentration of CO(g) in the equilibrium system, let's analyze each option:

a) Decreasing temperature: According to Le Chatelier's principle, decreasing the temperature of an equilibrium system will shift the reaction in the direction that generates heat. In this case, the reaction is exothermic (indicated by the negative enthalpy change). Therefore, decreasing the temperature would shift the equilibrium towards the side with fewer gaseous molecules, which is away from CO(g). As a result, the concentration of CO(g) would decrease, not increase. So, option a) is not correct.

b) Adding some CH3OH: Adding a reactant to an equilibrium system will shift the reaction in the direction that consumes it. In this case, adding CH3OH will shift the equilibrium towards the product side, consuming some CO(g). Therefore, the concentration of CO(g) would decrease, not increase. So, option b) is not correct.

c) Both b and d: According to Le Chatelier's principle, both adding a reactant (option b) and decreasing the volume of the container (option d) would shift the equilibrium towards the product side. As mentioned in option b, adding CH3OH would consume some CO(g), reducing its concentration. On the other hand, decreasing the volume of the container increases the pressure, which favors the side with fewer gaseous molecules. As CO(g) is the only gaseous species on the reactant side, decreasing the volume would shift the equilibrium towards the reactant side, increasing the concentration of CO(g). Therefore, option c) is correct.

d) Decreasing the volume of the container: As discussed above, decreasing the volume of the container increases the pressure, favoring the side with fewer gaseous molecules. Since CO(g) is the only gaseous species on the reactant side, decreasing the volume would shift the equilibrium towards the reactant side, increasing the concentration of CO(g). Therefore, option d) is also correct.

e) Both a and b: Option e) is not correct because option a) (decreasing temperature) would decrease the concentration of CO(g), as explained earlier.

So, the correct answer is c) both b and d, as both adding some CH3OH and decreasing the volume of the container would result in an increase in the concentration of CO(g) in the equilibrium system.

In "real" day to day talk (not chemical talk), Le Chatelier's Principle says that is we do something to a system at equilibrium, the system tries to undo what we've done. SO, as an example, adding H2 and the system will try to undo that. How can it do that. It can use up H2. How can it do that? By making more H2 and CO react to form more CH3OH. What about adding heat. The system will try to undo that; how can it do that? It will try to use the extra heat that we've added and it can do that by making the reaction shift to the left. As a result, CH3OH will be decreased, T will be decreased, H2 and CO will be increased. Using these two examples you should be able to answer the question. One point not emphasized above is about volume changes. If we decrease the volume of the container we must have increased the pressure (remember Boyle's Law). So the system will try to undo that by decreasing the pressure. It does that by shifting to the side with the smaller number of moles. In the above example that means a shift to the right (3 moles on the left and two on the right)