the reaction CO + 2H2<----->CH3OH has a delta H -18 kJ. how will the amount of CH3OH be affected by the following.
a) removing H2(g)
b) decreasing the volume of the container.
This is a question on Le Chatelier's Principle. Basically that principle tells us that a system in equilibrium will try to undo what we do to it. That isn't what you read in a text but it is one way to look at it.
So if we ADD H2 to the system it will try to undo that. How can it do that? By using up the added H2 of course. So the reaction will shift to the right (producing more CH3OH at the expense of H2 and CO). Increasing pressure will shift it to the side with FEWER moles. Take this information and answer the questions.
To determine how the amount of CH3OH will be affected by these changes, we need to consider the Le Chatelier's principle. According to this principle, a system at equilibrium will readjust itself to counteract any changes imposed on it.
a) Removing H2(g):
By removing H2(g) from the reaction mixture, we are essentially reducing the concentration of one of the reactants. According to Le Chatelier's principle, the system will respond by shifting the equilibrium position in the direction that compensates for the loss of H2(g). In this case, the reaction will shift to the left to restore the balance, resulting in a decrease in the amount of CH3OH.
b) Decreasing the volume of the container:
When the volume of the container is decreased, the pressure of the system increases due to the reduction in space for the gas molecules to move. According to Le Chatelier's principle, the system will respond by shifting the equilibrium position in the direction that reduces the number of gas molecules. In this case, the reaction will shift to the right to decrease the number of gas molecules by producing more CH3OH, resulting in an increase in the amount of CH3OH.
In summary, removing H2(g) from the reaction mixture will decrease the amount of CH3OH, while decreasing the volume of the container will increase the amount of CH3OH.