---
title: "An engineer studying the industrial synthesis of methanol from carbon monoxide and hydrogen gas according to the equation  \\[ \\text{CO}(g) + 2\\,\\text{H}_2(g) \\rightleftharpoons \\text{CH}_3\\text{OH}(g) \\]  plots the standard Gibbs free energy change, \\(\\Delta G^\\circ\\), as a function of absolute temperature, \\(T\\), as shown in the graph below.  Based on the graph, which of the following correctly identifies the signs of \\(\\Delta H^\\circ\\) and \\(\\Delta S^\\circ\\) for the forward reaction, and provides the correct thermodynamic justification?"
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url: "https://nerd-notes.com/ubq/120357/"
date_modified: "2026-08-23T04:23:22+00:00"
---

# An engineer studying the industrial synthesis of methanol from carbon monoxide and hydrogen gas according to the equation

\[ \text{CO}(g) + 2\,\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g) \]

plots the standard Gibbs free energy change, \(\Delta G^\circ\), as a function of absolute temperature, \(T\), as shown in the graph below.

Based on the graph, which of the following correctly identifies the signs of \(\Delta H^\circ\) and \(\Delta S^\circ\) for the forward reaction, and provides the correct thermodynamic justification?

An engineer studying the industrial synthesis of methanol from carbon monoxide and hydrogen gas according to the equation

\[ \text{CO}(g) + 2\,\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g) \]

plots the standard Gibbs free energy change, \(\Delta G^\circ\), as a function of absolute temperature, \(T\), as shown in the graph below.

Based on the graph, which of the following correctly identifies the signs of \(\Delta H^\circ\) and \(\Delta S^\circ\) for the forward reaction, and provides the correct thermodynamic justification?

![A line graph displaying standard Gibbs free energy change versus absolute temperature on a Cartesian coordinate plane. The horizontal axis is labeled \(T\text{ (K)}\) with tick marks at \(0\), \(200\), \(400\), \(600\), and \(800\). The vertical axis is labeled \(\Delta G^\circ\text{ (kJ/mol)}\) with tick marks at \(-100\), \(-50\), \(0\), \(50\), and \(100\). A horizontal dashed reference line extends across the plot at \(\Delta G^\circ = 0\text{ kJ/mol}\). A single solid black line is plotted with a constant positive slope, starting at a \(y\)-intercept of \(-90\text{ kJ/mol}\) at \(T = 0\text{ K}\), crossing the horizontal dashed axis at \((400, 0)\), and continuing upward to \((800, 90)\). No other curves, data points, shaded regions, labels, text, or annotations appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787459002-dAu7Tr.jpg)

- **A.** \(\Delta H^\circ > 0\) and \(\Delta S^\circ > 0\), because \(\Delta G^\circ\) increases with temperature, indicating that the reaction becomes thermodynamically favorable only at elevated temperatures.
- **B.** \(\Delta H^\circ < 0\) and \(\Delta S^\circ > 0\), because the negative \(y\)-intercept corresponds to \(\Delta H^\circ < 0\) and the positive slope directly indicates that \(\Delta S^\circ > 0\).
- **C.** \(\Delta H^\circ < 0\) and \(\Delta S^\circ < 0\), because the negative \(y\)-intercept represents \(\Delta H^\circ\) and the positive slope represents \(-\Delta S^\circ\), meaning \(\Delta S^\circ\) must be negative.
- **D.** \(\Delta H^\circ > 0\) and \(\Delta S^\circ < 0\), because the reaction is nonspontaneous (\(\Delta G^\circ > 0\)) above \(400\text{ K}\), which requires an endothermic enthalpy change.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120357/*