CH4 (g)+ H2O (g) --->/<---- 3H2 + CO
Use Le Chateliers Principle to explain what will happen to the number of moles of hydrogen at equilibrium when
a) CO is removed
b) Water Vapor is removed
c) Methane is added
d) CO is added
e) Pressure is increased by adding argon gas, but the partial pressure of the reactants and products do not change.
f) The pressure is decreased by opening a valve to a second reaction vessel, which lowers the partial pressures of the reactants and products.
g) The temperature is increased.
The principle says if a change occurs at equilibrium, the reaction will shift to "undo" the change.
a. CO is removed, the reaction will go to the right to make more CO.
b) The reaction will shift to the left to make more H2O (g).
g) Not indicated, is the heat of reaction, however, this reaction is endothermic (it takes heat to change methane and steam to make hydrogen gas). So if you increase temp, it will shift to the right. Increasing temp is adding heat.
a) When CO is removed, according to Le Chatelier's Principle, the equilibrium will shift to the right to replace the CO that was removed. This means that more H2 and CO will react to form additional moles of H2. Therefore, the number of moles of hydrogen at equilibrium will increase.
b) When water vapor is removed, the equilibrium will shift to the left to replace the water vapor that was removed. This means that some of the H2 and CO will react to form more water vapor. Since water vapor is on the left side of the equation, the reaction will consume some of the hydrogen, resulting in a decrease in the number of moles of hydrogen at equilibrium.
c) When methane is added, the equilibrium will shift to the right to consume the additional methane. This means that more H2 and CO will react to form additional moles of H2. Therefore, the number of moles of hydrogen at equilibrium will increase.
d) When CO is added, the equilibrium will shift to the left to consume the additional CO. This means that some of the H2 will react with the added CO to form more water vapor. Since water vapor is on the left side of the equation, the reaction will consume some of the hydrogen, resulting in a decrease in the number of moles of hydrogen at equilibrium.
e) When pressure is increased by adding argon gas, it does not directly affect the equilibrium position or the number of moles of hydrogen. According to Le Chatelier's Principle, changing the total pressure of a system does not affect the equilibrium position if the partial pressures of the reactants and products remain constant.
f) When the pressure is decreased by opening a valve to a second reaction vessel, it will lower the partial pressures of the reactants and products. According to Le Chatelier's Principle, the equilibrium will shift in the direction that produces more moles of gas. Since there are more moles of gas on the left side of the equation (CH4 and H2O) compared to the right side (H2 and CO), the equilibrium will shift to the right to produce more moles of gas, including hydrogen. Therefore, the number of moles of hydrogen at equilibrium will increase.
g) When the temperature is increased, the equilibrium will shift in the direction that absorbs heat. In the given reaction, the forward reaction is endothermic (heat is a reactant) as indicated by the positive delta H value. Thus, increasing the temperature will shift the equilibrium to the right to consume some of the excess heat. Since the reaction proceeds in the direction that produces more moles of gas, the number of moles of hydrogen at equilibrium will increase.
To use Le Chatelier's principle to explain what happens to the number of moles of hydrogen gas at equilibrium in different scenarios, we need to understand how the given reaction is affected by changes in reactants, products, temperature, and pressure.
The given reaction is a synthesis reaction, where methane gas (CH4) and water vapor (H2O) combine to produce hydrogen gas (H2) and carbon monoxide gas (CO). The balanced reaction equation is as follows:
CH4 (g) + H2O (g) ⇌ 3H2 (g) + CO (g)
Now, let's analyze each scenario:
A) When CO is removed:
Removing carbon monoxide (CO) from the equilibrium mixture will cause the reaction to shift to the right to replace the lost CO. According to Le Chatelier's principle, by removing one of the products, the equilibrium will favor the forward reaction to offset the change. As a result, the number of moles of hydrogen gas (H2) will increase at equilibrium.
B) When water vapor is removed:
Removing water vapor (H2O) from the equilibrium mixture will also cause the reaction to shift to the right to replenish the lost water vapor. This shift favors the forward reaction, leading to an increase in the number of moles of hydrogen gas (H2) at equilibrium.
C) When methane is added:
Adding methane (CH4) to the equilibrium mixture will increase the concentration of reactants. As per Le Chatelier's principle, the equilibrium will favor the reverse reaction to consume the excess methane. Consequently, the number of moles of hydrogen gas (H2) will decrease at equilibrium.
D) When CO is added:
Adding carbon monoxide (CO) to the equilibrium mixture will increase the concentration of reactants. Following Le Chatelier's principle, the equilibrium will move in the reverse direction to compensate for the increased CO. This shift favors the reverse reaction, reducing the number of moles of hydrogen gas (H2) at equilibrium.
E) When pressure is increased by adding argon gas (with no change in partial pressures):
The increase in pressure due to the addition of inert argon gas will not affect the partial pressures of reactants or products. Le Chatelier's principle states that changes in overall pressure only affect the equilibrium if they change the partial pressures of the reactants or products. Consequently, the number of moles of hydrogen gas (H2) remains unchanged at equilibrium.
F) When pressure is decreased by opening a valve to a second reaction vessel:
Reducing the pressure by opening a valve decreases the overall pressure of the system. As a result, according to Le Chatelier's principle, the reaction will shift to the side with the greater number of moles of gas molecules. In this case, since there are more moles of gas on the reactant side (CH4 + H2O), the equilibrium will shift to the left, causing a decrease in the number of moles of hydrogen gas (H2) at equilibrium.
G) When the temperature is increased:
Increasing the temperature affects the equilibrium of this reaction. According to Le Chatelier's principle, if the reaction is exothermic (releases heat), increasing the temperature will shift the equilibrium in the reverse (left) direction to consume the excess heat. On the other hand, if the reaction is endothermic (absorbs heat), increasing the temperature will shift the equilibrium in the forward (right) direction to counteract the increased temperature. Without knowing whether the reaction is endothermic or exothermic, it is not possible to determine the effect on the number of moles of hydrogen gas at equilibrium.
In summary:
a) The number of moles of hydrogen gas (H2) increases.
b) The number of moles of hydrogen gas (H2) increases.
c) The number of moles of hydrogen gas (H2) decreases.
d) The number of moles of hydrogen gas (H2) decreases.
e) The number of moles of hydrogen gas (H2) remains unchanged.
f) The number of moles of hydrogen gas (H2) decreases.
g) The effect is uncertain without knowing the direction of the reaction (endothermic or exothermic).