- The reaction has 2 moles of gas on the reactant side (\(CH_4 + H_2O\)) and 4 moles of gas on the product side (\(CO + 3H_2\)).
- According to Le Chatelier's Principle, increasing the pressure of a system at equilibrium shifts the reaction towards the side with fewer moles of gas to reduce the pressure.
- Since the reactant side has fewer moles (2) than the product side (4), an increase in pressure shifts the equilibrium to the left.
- Therefore, the equilibrium shifts to favor the reverse reaction, which corresponds to option \(\boxed{b}\).
Explanation
- Analyzing the Reaction and Stoichiometry
The chemical reaction provided is given by the balanced chemical equation:
\[CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g)\]
To understand how the system reacts to changes in pressure, we first identify the total number of moles of gas present on both sides of the equation.
On the reactant side (left), we have one mole of \(CH_4(g)\) and one mole of \(H_2O(g)\), resulting in:
\[1 + 1 = 2 \text{ moles of gas}\]
On the product side (right), we have one mole of \(CO(g)\) and three moles of \(H_2(g)\), resulting in:
\[1 + 3 = 4 \text{ moles of gas}\]
We observe that the reactant side has 2 moles of gas and the product side has 4 moles of gas.
- Understanding Le Chatelier's Principle for Pressure
Le Chatelier's Principle provides a framework for predicting how a system at equilibrium will respond to changes in conditions such as pressure, temperature, or concentration. Specifically, regarding pressure, Le Chatelier's Principle states that when the pressure of a system at equilibrium is increased, the system will shift its equilibrium position to counteract the change. It does this by favoring the side of the reaction that results in fewer moles of gas, thereby reducing the total pressure within the vessel. Conversely, if the pressure is decreased, the system shifts toward the side with more moles of gas to increase the pressure.
- Predicting the Shift in Equilibrium
We have established that the reactant side consists of 2 moles of gas, while the product side consists of 4 moles of gas. When we increase the total pressure of the system, the system seeks to minimize this stress by moving towards the direction that generates fewer moles of gas. Since the reactant side has fewer moles of gas (2 moles) compared to the product side (4 moles), the increase in pressure causes the equilibrium to shift to the left, favoring the reverse reaction (the formation of \(CH_4\) and \(H_2O\)).
- Conclusion and Final Answer
Based on the analysis, increasing the pressure favors the direction with fewer moles of gas. Because there are 2 moles of gas on the reactant side and 4 moles of gas on the product side, the equilibrium will shift to the left to favor the reverse reaction. This matches option b. Thus, the correct answer is that the equilibrium will shift to the left to favor the reverse reaction. \(\boxed{b}\)
Examples
Le Chatelier's Principle is essential in industrial chemistry, such as the Haber-Bosch process for synthesizing ammonia (\(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)). By understanding how pressure affects the equilibrium, engineers can optimize reaction conditions to maximize the yield of desired products. For example, by applying high pressure to the synthesis of ammonia, where 4 moles of gaseous reactants produce 2 moles of gaseous product, the equilibrium shifts to the right, increasing the efficiency of production. Similar principles are used in various chemical manufacturing processes to ensure efficient use of resources.