Use the following equation:
CO(g) + 2H2 (g)= CH3OH(g) delta H = -21.7 kcal
Which direction will the reaction proceed after some of the methanol vapor is condensed and removed from the reaction vessel?
please help im getting so confused on the directions of the reactions
This is all about Le Chatelier's Principle and it's easy if you just remember a couple of things.
1. The principle tells us that when a system at equilibrium is disturbed, it will react to undo what we did to it.
a. An increase in pressure shifts it to the side with fewer GAS moles.
b. An increase in reactants or products shifts the reaction AWAY from what you have added.
c. An increase in T is treated as in b.
d. Adding an inert gas has no effect.
So if we remove CH3OH the reaction tries to replace what we've removed. How can it do that. Easy. Make more so the reaction shifts to the right.
What if we add H2 or CO. Now there is too much CO or H2, the reaction must rreact to get rid of some of it. How can it do that. Easy. The reaction to the right so as to use up the added CO or added H2. In this case more CH3OH is produced the same as if we remove some of the CH3OH. Students get so confused about this one concept in chemistry. But you can know what is what just by following the logic above.
To determine the direction in which the reaction will proceed after some of the methanol vapor is condensed and removed, you need to consider Le Chatelier's principle.
In this case, when some of the methanol vapor is removed, it results in a decrease in the concentration of CH3OH(g) on the right side of the equation. According to Le Chatelier's principle, the reaction will shift in the direction that replenishes the missing reactant. In other words, the reaction will proceed in the forward direction to produce more CH3OH(g) until the equilibrium is reestablished.
Therefore, after some of the methanol vapor is condensed and removed from the reaction vessel, the reaction will proceed in the forward direction to produce more methanol gas.
To determine the direction in which the reaction will proceed after removing some of the methanol vapor, we need to consider Le Chatelier's principle. According to this principle, when a change is made to a system at equilibrium, the system will respond in such a way as to counteract that change.
In this case, by condensing and removing methanol vapor, we are effectively reducing the amount of CH3OH(g) in the reaction vessel. To restore equilibrium, the reaction will shift in the direction that produces more methanol.
To identify the direction of the reaction, we can examine the stoichiometric coefficients of the reactants and products.
The equation you provided is:
CO(g) + 2H2(g) = CH3OH(g)
Based on the stoichiometry, we see that for each CO molecule and 2 H2 molecules, one CH3OH molecule is formed. Therefore, the formation of CH3OH is favored when the reactant concentrations decrease.
Since we are removing some of the methanol vapor, the concentration of CH3OH is decreasing, which means the reaction will shift in the forward direction (from left to right) to restore equilibrium. This will result in more CO and H2 reacting to produce additional CH3OH.
In summary, after removing some of the methanol vapor, the reaction will proceed in the forward direction (from left to right) to replenish the reduced concentration of methanol.