Consider the reaction
3Fe(s) + 4H2O(g) �¨ 4H2(g) + Fe3O4(s).
If the volume of the container is reduced,
1. more Fe(s) is produced.
2. no change occurs.
3. more H2O(g) is produced.
4. more H2(g) is produced.
5. the equilibrium constant increases.
Le Chatelier's Principle says that a system in equilibrium will try to undo what we do to it. When volume is reduced that means pressure must be increased so the system will shift to the side with the fewer mols of gas. In this reaction there are 4 mols gas on the left and 4 mols gas on the right; therefore, no change occurs.
We want word equestion
When the volume of the container is reduced, it affects the reaction equilibrium. In this reaction, 3Fe(s) + 4H2O(g) �¨ 4H2(g) + Fe3O4(s), the reaction shifts to consume the excess product in order to regain the equilibrium state.
Since the reaction is consuming the excess product, more reactant Fe(s) will produce to compensate for the decrease in volume. Thus, option 1 is correct: more Fe(s) is produced.
Option 2 is not the correct answer because a change is occurring due to the decrease in volume.
Option 3 is not the correct answer because the reaction is not producing more H2O(g) as a result of the volume decrease.
Option 4 is not the correct answer because the reaction is consuming H2O(g) to produce H2(g) in order to regain equilibrium, not producing more H2(g).
Option 5 is not the correct answereither because the equilibrium constant is a constant value for a particular reaction at a given temperature, and it does not change with volume changes. The equilibrium constant is determined by the temperature and the relative concentrations of the reactants and products.
To determine the effect of reducing the volume of the container on the given reaction, we need to consider the principles of Le Chatelier's principle.
Le Chatelier's principle states that when a system at equilibrium is subjected to a change in conditions, the system will shift to counteract the change and restore equilibrium.
In this case, if the volume of the container is reduced, the reaction will experience an increase in pressure. Let's analyze each option based on this information:
1. More Fe(s) is produced: The reaction involves the formation of Fe(s), but since the increase in pressure doesn't affect the amount of Fe(s) present, this option is incorrect.
2. No change occurs: It is unlikely that no change will occur. According to Le Chatelier's principle, a shift in the equilibrium position will happen to restore equilibrium.
3. More H2O(g) is produced: Since the increase in pressure favors the side with fewer moles of gas, which in this case is the reactant side (H2O(g)), it is expected that the system will shift to produce more H2O(g). Therefore, this option is correct.
4. More H2(g) is produced: The formation of H2(g) is favored when the pressure increases since it corresponds to an increase in moles of gas. This option is incorrect.
5. The equilibrium constant increases: The equilibrium constant is determined by the concentrations of the reactants and products, not by changes in pressure or volume. Therefore, this option is incorrect.
In conclusion, option 3, "more H2O(g) is produced," is the correct answer.