CO(g) + 2 H2(g) ⇄ CH3OH(g) DH < 0 The synthesis of CH3OH(g) from CO(g) and H2(g) is represented by t — Equilibrium Chemistry Question

STEPS:

1. Understand the initial state of the system: The chemical reaction is $\text{CO}(g) + 2\text{ H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)$. The gas mixture is at dynamic equilibrium in a $2.0\text{ L}$ vessel at a constant temperature of $483\text{ K}$ and exerts an initial total pressure of $1.2\text{ atm}$.
2. Calculate the instantaneous effect of reducing the volume (Boyle's Law): The volume of the rigid reaction vessel is reduced from $2.0\text{ L}$ to $1.0\text{ L}$. According to Boyle's Law ($PV = \text{constant}$ at constant temperature and moles), halving the volume of a gas mixture immediately doubles the partial pressures of all individual gases. If no chemical shift were to occur, the total pressure of the system would instantly double from $1.2\text{ atm}$ to exactly:
$$1.2\text{ atm} \times 2 = \mathbf{2.4\text{ atm}} \quad$$
3. Apply Le Chatelier's Principle to the pressure increase: Halving the volume increases the overall concentration and total pressure of the system, creating a pressure stress. According to Le Chatelier's Principle, a system at equilibrium subjected to an increase in pressure will shift its equilibrium position in the direction that minimizes this pressure by producing fewer moles of gas.
4. Compare the moles of gaseous reactants and products:
* Reactants: $1\text{ mole of CO}(g) + 2\text{ moles of H}_2(g) = \mathbf{3\text{ moles of gas}}$
* Products: $\mathbf{1\text{ mole of CH}_3\text{OH}(g)}$
Since the forward reaction converts 3 moles of gas into 1 mole of gas, a shift to the right decreases the total number of gas molecules in the vessel. Therefore, the equilibrium will shift to the right (the forward direction).
5. Determine the final equilibrium pressure:
* As the system shifts forward to produce more $\text{CH}_3\text{OH}(g)$, the total number of gaseous particles in the vessel decreases, which lowers the total pressure below the instantaneous peak of $2.4\text{ atm}$.
* However, Le Chatelier's shift can only *partially* relieve the applied pressure stress; it cannot completely counteract the effect of halving the volume. Thus, the final equilibrium pressure must remain higher than the original equilibrium pressure of $1.2\text{ atm}$.
* Combining these bounds, the final total pressure of the system will be greater than 1.2 atm but less than 2.4 atm, which is Option B.

*

WHY_OTHERS_WRONG:

• A is incorrect: The total pressure cannot be less than $1.2\text{ atm}$. Halving the container volume packs the molecules into a much tighter space, which increases the frequency of collisions. The chemical shift can only partially mitigate this increase, so the final pressure must remain higher than the initial pressure.
• C is incorrect: A final pressure of exactly $2.4\text{ atm}$ would only occur if the system did not undergo any chemical shift. Since the reactants and products have unequal moles of gas (3 moles vs 1 mole), a forward shift is guaranteed, reducing the pressure below $2.4\text{ atm}$.
• D is incorrect: A final pressure greater than $2.4\text{ atm}$ would require the reaction to shift to the left (toward the reactants) to produce *more* gas molecules. This would increase the total pressure inside the container, violating Le Chatelier's Principle by amplifying the applied stress rather than relieving it.