Final Answer:
The reaction that best accounts for the observed increase in pressure at 127°C in vessel 1 compared to vessel 2 is C) CO(g) + 2H₂(g) → CH₃OH(g).
Step-by-step explanation:
At 127°C, the reaction (CO(g) + 2H₂(g) → CH₃OH(g)) involves the formation of methanol (CH₃OH) from carbon monoxide (CO) and hydrogen (H₂). This reaction is exothermic and leads to the production of a greater number of moles of gas on the right side of the equation compared to the left side. The increase in temperature facilitates the forward reaction, favoring the production of methanol and contributing to a higher pressure in vessel 1. The other options involve either fewer moles of gas or endothermic reactions, making them less likely to result in an increased pressure at the given temperature.
To delve into the specifics, the stoichiometry of the reaction is critical. For every one mole of CO and two moles of H₂, one mole of CH₃OH is produced. The shift towards the product side at an elevated temperature enhances the pressure in vessel 1. This aligns with Le Chatelier's Principle, stating that a system at equilibrium, when subjected to changes such as temperature, will adjust to counteract those changes.
In summary, the reaction (CO(g) + 2H₂(g) → CH₃OH(g)) is the most plausible explanation for the observed increase in pressure at 127°C in vessel 1. The chemical reaction's thermodynamics, particularly its exothermic nature and the resultant increase in gas moles, support the conclusion.