Question

Data given for the following reaction is as follows: 
FeO(s) + C(graphite) \(\rightarrow\) Fe(s) + CO(g) 
The minimum temperature in K at which the reaction becomes spontaneous is _________. (Integer answer) 

 

Hint:

When calculating the spontaneity crossover temperature, always check the signs of \(\Delta H\) and \(\Delta S\). If both are positive, the reaction becomes spontaneous at high temperatures (\(T>\Delta H / \Delta S\)). If both are negative, it's spontaneous at low temperatures (\(T<\Delta H / \Delta S\)). Also, be very careful with units, especially converting kJ to J for enthalpy.

Answer 1

Correct Answer: 964

Step 1: Understanding the Question:
We need to find the temperature at which the given reaction becomes spontaneous. A reaction is spontaneous when the Gibbs free energy change (\(\Delta G\)) is negative. The minimum temperature for spontaneity occurs at the point where \(\Delta G = 0\).
Step 2: Key Formula:
The Gibbs free energy change is related to enthalpy and entropy by the equation:
\[ \Delta G = \Delta H - T\Delta S \] At equilibrium (the crossover point for spontaneity), \(\Delta G = 0\), so \(T = \frac{\Delta H}{\Delta S}\).
Step 3: Calculate \(\Delta H^\circ\) and \(\Delta S^\circ\) for the Reaction:
We use the formula: \(\Delta X^\circ_{rxn} = \sum X^\circ_{products} - \sum X^\circ_{reactants}\)
- Enthalpy Change (\(\Delta H^\circ\)):
\[ \Delta H^\circ = [\Delta H_f^\circ(\text{Fe}) + \Delta H_f^\circ(\text{CO})] - [\Delta H_f^\circ(\text{FeO}) + \Delta H_f^\circ(\text{C})] \] \[ \Delta H^\circ = [0 + (-110.5)] - [(-266.3) + 0] = 155.8 \, \text{kJ/mol} \] - Entropy Change (\(\Delta S^\circ\)):
\[ \Delta S^\circ = [S^\circ(\text{Fe}) + S^\circ(\text{CO})] - [S^\circ(\text{FeO}) + S^\circ(\text{C})] \] \[ \Delta S^\circ = [27.28 + 197.6] - [57.49 + 5.74] = 224.88 - 63.23 = 161.65 \, \text{J/mol K} \] Step 4: Calculate the Temperature:
We need to use consistent units. Let's convert \(\Delta H^\circ\) to J/mol.
\[ \Delta H^\circ = 155.8 \, \text{kJ/mol} = 155800 \, \text{J/mol} \] Now, calculate the temperature T.
\[ T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{155800 \, \text{J/mol}}{161.65 \, \text{J/mol K}} \approx 963.81 \, \text{K} \] Since both \(\Delta H^\circ\) and \(\Delta S^\circ\) are positive, the reaction is non-spontaneous at low temperatures and becomes spontaneous at temperatures above this value.
Rounding to the nearest integer, the minimum temperature is 964 K.