Mr Cole Chemistry

Gibbs Free Energy Graphs Questions

Graphs & Feasibility

Thermodynamics

1. Below is the graph showing the relationship between Temperature and Gibbs Free Energy for the reaction between iron (III) oxide and carbon, producing carbon dioxide and iron.

Graph of Gibbs Free Energy vs Temperature for Iron extraction
a) Write an equation for the reaction.
$2Fe_2O_3(s) + 3C(s) \rightarrow 4Fe(s) + 3CO_2(g)$
b) Use the shape of the graph to deduce whether the reaction is exothermic or endothermic, and whether the reaction has a positive or negative entropy change.
Analysis of Gradient

The graph has a negative gradient. Since Gradient = $-\Delta S$, a negative gradient means $-\Delta S$ is negative, so $\Delta S$ is positive.

Feasibility Trend

$\Delta G$ becomes more negative (more feasible) as temperature increases. This is characteristic of an endothermic reaction with increasing entropy.

Endothermic with Positive Entropy change.
c) Use the graph to determine the temperature range at which this reaction is feasible.
Identifying the intercept

Feasibility occurs when $\Delta G \le 0$. On the graph, the line crosses $\Delta G = 0$ (the x-axis) at approximately 835 K.

Feasible at 835 K (± 5 K) and above.
d) Use the graph to determine the enthalpy change of the reaction.
Using the Y-Intercept

Based on the equation $y = mx + c$ corresponding to $\Delta G = -\Delta S(T) + \Delta H$.

The Y-intercept (when $T=0$) represents $\Delta H$.

$\Delta H = +470 \text{ kJ mol}^{-1}$
e) Use the graph to determine the entropy change of the reaction.
Calculate Gradient

Gradient = $\frac{\Delta y}{\Delta x} = \frac{0 – 470}{835 – 0} = -0.563 \text{ kJ K}^{-1} \text{mol}^{-1}$

Convert to Entropy

Gradient = $-\Delta S$

$-0.563 = -\Delta S \Rightarrow \Delta S = +0.563 \text{ kJ K}^{-1} \text{mol}^{-1}$

$\Delta S = +563 \text{ J K}^{-1} \text{mol}^{-1}$

2. The table below shows the effect of temperature on the Gibbs Free Energy of a reaction.

Temp / K Gibbs Free Energy / kJ mol-1
0185
1500138
3000102
450071.5
600030.0
7500-8.50
9000-40.0
10500-87.5
12000-120
13500-157

Draw a graph of this data and use the graph to determine the following:

a) The range of temperatures that this reaction is feasible at.
Analyze Graph Intercept

Plotting the points allows you to find the X-intercept (where $\Delta G = 0$).

The line crosses the X-axis at approximately 7200 K.

Feasible at 7200 K and above.
b) The entropy change of the reaction.
Calculate Gradient

Using the intercept points: $(0, 185)$ and $(7200, 0)$.

Gradient = $\frac{0 – 185}{7200 – 0} = -0.025 \text{ kJ K}^{-1} \text{mol}^{-1}$

Determine Entropy

Gradient = $-\Delta S$, so $\Delta S = 0.025 \text{ kJ K}^{-1} \text{mol}^{-1}$

$\Delta S = 25 \text{ J K}^{-1} \text{mol}^{-1}$
c) The enthalpy change of the reaction.
Y-Intercept

The value of Gibbs Free Energy at 0 K represents the Enthalpy change.

$\Delta H = +185 \text{ kJ mol}^{-1}$

3. The reaction between carbon dioxide and hydrogen produces methanol and water. The effect of temperature on the Gibbs Free Energy is shown in the graph.

Graph of Gibbs Energy for Methanol production

Use the graph to determine the following:

a) The temperature range at which the reaction is feasible.
X-Intercept Analysis

The graph crosses the X-axis (where $\Delta G = 0$) at approximately 955 K.

Since the gradient is negative, the reaction is feasible at temperatures higher than this intercept.

Feasible at 955 K and above.
b) The enthalpy change of the reaction.
Y-Intercept

The graph starts at 205 on the Y-axis.

$\Delta H = +205 \text{ kJ mol}^{-1}$
c) The entropy change of the reaction.
Calculate Gradient

Gradient = $\frac{0 – 205}{955 – 0} = -0.215 \text{ kJ K}^{-1} \text{mol}^{-1}$

Determine Entropy

Since Gradient = $-\Delta S$, then $\Delta S = 0.215 \text{ kJ K}^{-1} \text{mol}^{-1}$

$\Delta S = +215 \text{ J K}^{-1} \text{mol}^{-1}$
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