Remember, LeChatlier's Principle says that if a stress is applied to a reaction the reaction will naturally shift away from the applied stress to remove it. This will continue until a new equilibrium is established.
For your problem, in order increase the concentration of HOH, a stress would need to be applied to the product side of the reaction. Such would cause the reaction to shift left and, as a result, increase the HOH concentration. This would continue until the stress is removed and a new equilibrium would be established.
'D' choice is your best answer. Decreasing Volume of container increases Pressure (Boyles Law). The natural result is for the reaction to shift to the side of the equation having the LOWER moles of gas. In this case a left shift which increases the HOH concentration.
A => Increasing temp puts the stress on the reactant side and shifts reaction right. In future problems, look for the energy number in the equation. If temp is increased assume more weight is being added to that side causing the rxn to shift away from the energy number. If temperature is decreased, the opposite is true.
B => Decreasing Pressure causes a gas phase reaction to shift to the side have the higher moles of gas. Increasing pressure shifts reaction to lower moles of gas side.
C => Removing CO from the product side lightens the load on the product side and more reactants will react to replace what's been removed. In this case that would be a shift right to replace lost CO on the product side. This will continue until the concentration of CO has increased sufficiently to establish a new equilibrium. Hope this helps.
7. For the equilibrium system below, which of the following would increase the concentration of π»2π(π)?
πΆ(π ) + π»2π(π) + 131ππ½ β πΆπ(π) + π»2(π)
a. Increased temperature
b. Decreased pressure
c. Decreased [CO]
d. Decreased container volume
1 answer
1 answer