Hydrogen for use in production of ammonia is shown in the reaction:
CH4 (g)+ H2O (g) ⎯⎯⎯⎯⎯⎯→ NiCatalyst,750C CO (g) + 3 H2 (g)
What will happen to above reaction mixture at equilibrium if H2O (g) is removed?
Le Chatelier's Principle tells us that when we do something to a system in equilibrium that the system will try to undo whatever we do to it. So if we remove H2O(g) from the reaction, it will try to add water back; i.e., the reaction will shift to the left which means the CO and H2 will be reduced.
If H2O (g) is removed from the reaction mixture, according to Le Chatelier's principle, the equilibrium position will shift to counteract the change. In this case, the reaction favors the formation of more H2O (g) because it was removed. Therefore, to compensate for the loss of H2O (g), the reaction will shift towards the reactants' side. Consequently, more CH4 (g) and CO (g) will be formed, while the concentration of H2 (g) will decrease.
In this reaction, methane (CH4) reacts with water (H2O) in the presence of a nickel catalyst at a temperature of 750°C to produce carbon monoxide (CO) and hydrogen gas (H2).
The reaction can be represented as follows:
CH4 (g) + H2O (g) → Ni Catalyst, 750°C CO (g) + 3 H2 (g)
Now, let's consider the effect of removing water (H2O) from the reaction mixture at equilibrium.
According to Le Chatelier's principle, when a system at equilibrium is disturbed, it will respond in such a way as to minimize the effect of the disturbance and restore equilibrium.
In this case, by removing water (H2O), we are essentially decreasing the concentration of one of the reactants. As a result, the reaction will try to compensate for this change by shifting in the direction that produces more water.
In the given reaction, water is consumed in order to produce carbon monoxide and hydrogen gas. Therefore, by removing water, the reaction will shift to the left to form more water (H2O) in order to restore equilibrium.
As a result, the concentrations of carbon monoxide (CO) and hydrogen gas (H2) will decrease, while the concentration of methane (CH4) will remain unaffected.
It's important to note that the reaction will continue to shift until a new equilibrium is reached, where the ratio of reactants and products is once again balanced.
In summary, removing water (H2O) from the reaction mixture at equilibrium will cause the reaction to shift to the left, resulting in a decrease in the concentrations of carbon monoxide (CO) and hydrogen gas (H2), while methane (CH4) will remain unaffected.