Data given for the following reaction is as follows. FeO(s) + C(graphite) - Fe(s) + CO(g). Substanc

Q: Why are the enthalpy values for Fe(s) and C(graphite) zero?

By thermodynamic convention, the standard enthalpy of formation for any element in its most stable reference state (like solid iron or graphite for carbon) is defined as exactly zero.

Q: Why do we set $\Delta G^\circ = 0$ to find the temperature?

Setting $\Delta G^\circ = 0$ gives us the equilibrium temperature. This is the exact tipping point. Any temperature above this (for an entropy-driven reaction) will make $\Delta G^\circ$ negative, triggering spontaneity.

Q: Why did we multiply $\Delta H^\circ$ by 1000?

This is the most common trap! Enthalpy is provided in kiloJoules (kJ) while entropy is in Joules (J). To mathematically operate on them together, they must be in the same base unit. Multiplying kJ by 1000 converts it to Joules.

Q: What does it mean for a reaction to be 'entropy-driven'?

When a reaction is endothermic ($\Delta H^\circ > 0$), it naturally resists happening. However, if it creates more disorder ($\Delta S^\circ > 0$), high temperatures will amplify this entropy term ($T\Delta S^\circ$), eventually overpowering the enthalpy barrier and driving the reaction forward.

Q: Could this reaction be spontaneous at room temperature?

No. At room temperature (around 298 K), the $T\Delta S^\circ$ term is too small to overcome the large positive $\Delta H^\circ$. The overall $\Delta G^\circ$ remains positive, meaning the reaction is non-spontaneous.

Question

Data given for the following reaction is as follows.
$	ext{FeO}(s) + 	ext{C}_{(	ext{graphite})} \longrightarrow 	ext{Fe}(s) + 	ext{CO}(g)$

\begin{array}{ccc} \hline 	ext{Substance} & \Delta H^\circ (	ext{kJ mol}^{-1}) & \Delta S^\circ (	ext{J mol}^{-1} 	ext{K}^{-1}) \ \hline 	ext{FeO}(s) & -266.3 & 57.49 \ 	ext{C}_{(	ext{graphite})} & 0 & 5.74 \ 	ext{Fe}(s) & 0 & 27.28 \ 	ext{CO}(g) & -110.5 & 197.6 \ \hline \end{array}

The minimum temperature in K at which the reaction becomes spontaneous is ........... (Integer answer)

Sigma Percentile · Accepted Answer

The universe is governed by a constant tug-of-war between two fundamental forces: the drive towards lower energy (enthalpy) and the drive towards maximum chaos (entropy).

When we are asked to find the minimum temperature at which a reaction becomes spontaneous, we are essentially looking for the exact moment when the chaos factor overpowers the energy barrier.

Let's embark on this thermodynamic journey and decode the math behind the magic!

### Decoding the Gibbs Free Energy Equation

The ultimate judge of spontaneity is the Gibbs Free Energy change, denoted by $\Delta G^\circ$.

The master equation that connects enthalpy, entropy, and temperature is:

$$ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ $$

For any process to occur spontaneously, the universe demands that $\Delta G^\circ$ must be strictly less than zero ($\Delta G^\circ < 0$).

However, to find the **threshold** temperature—the exact tipping point where the reaction transitions from non-spontaneous to spontaneous—we must find the state of perfect equilibrium. At equilibrium, the free energy change is exactly zero ($\Delta G^\circ = 0$).

Setting the equation to zero gives us our working formula:

$$ T_{eq} = \frac{\Delta H^\circ}{\Delta S^\circ} $$

### Step 1: Calculating the Enthalpy Change ($\Delta H^\circ$)

Our first mission is to find the total enthalpy change of the reaction. We do this by subtracting the total enthalpy of the reactants from the total enthalpy of the products.

$$ \Delta H^\circ_{	ext{rxn}} = \sum \Delta H^\circ_{f(	ext{products})} - \sum \Delta H^\circ_{f(	ext{reactants})} $$

Looking at our reaction, we have solid Iron ($	ext{Fe}$) and Carbon graphite ($	ext{C}$) involved. By thermodynamic convention, elements in their standard, most stable states have an enthalpy of formation of exactly zero.

Substituting the values from our data table:

$$ \Delta H^\circ_{	ext{rxn}} = [0 + (-110.5)] - [-266.3 + 0] $$

$$ \Delta H^\circ_{	ext{rxn}} = 155.8 	ext{ kJ mol}^{-1} $$

Notice that the result is positive. This means the reaction is **endothermic**; it requires an input of heat to proceed. From an enthalpy perspective, this reaction does not want to happen!

### Step 2: Calculating the Entropy Change ($\Delta S^\circ$)

Now, let's evaluate the chaos! We calculate the standard entropy change using the exact same principle: products minus reactants.

$$ \Delta S^\circ_{	ext{rxn}} = \sum \Delta S^\circ_{	ext{products}} - \sum \Delta S^\circ_{	ext{reactants}} $$

We carefully pull the entropy values from the table for each substance:

$$ \Delta S^\circ_{	ext{rxn}} = [27.28 + 197.6] - [57.49 + 5.74] $$

$$ \Delta S^\circ_{	ext{rxn}} = 224.88 - 63.23 $$

$$ \Delta S^\circ_{	ext{rxn}} = 161.65 	ext{ J K}^{-1} 	ext{mol}^{-1} $$

The entropy change is positive! This is fantastic news. It means the reaction creates more disorder, primarily because we are generating Carbon Monoxide ($	ext{CO}$) gas from solid reactants. The universe loves this increase in chaos.

### The Crucial Trap: Unit Consistency

Here is where countless students lose their marks. We have our $\Delta H^\circ$ and $\Delta S^\circ$, but look closely at their units!

Enthalpy is measured in **kiloJoules** ($	ext{kJ}$), while entropy is measured in standard **Joules** ($	ext{J}$). We cannot mathematically divide them as they are. We must bring them to the same base unit.

Let's convert the enthalpy into Joules by multiplying by $1000$:

$$ \Delta H^\circ_{	ext{rxn}} = 155.8 	imes 10^3 	ext{ J mol}^{-1} $$

### Finding the Threshold Temperature

With our units perfectly aligned, we are ready for the final execution. We substitute our values into the threshold temperature formula:

$$ T_{eq} = \frac{155.8 	imes 10^3}{161.65} $$

$$ T_{eq} = 963.81 	ext{ K} $$

The question asks for the nearest integer. Rounding $963.81$ gives us our final answer: **$964 	ext{ K}$**.

### The Physical Significance of the Result

What does this number actually mean?

Because both our enthalpy ($\Delta H^\circ$) and entropy ($\Delta S^\circ$) are positive, this is what chemists call an **entropy-driven reaction**.

At low temperatures, the $T\Delta S^\circ$ term is too weak to overcome the massive $155.8 	ext{ kJ}$ energy barrier. The reaction sits dormant.

But as we crank up the heat, the temperature multiplier ($T$) amplifies the chaos term. The moment the temperature crosses $964 	ext{ K}$, the $T\Delta S^\circ$ term becomes larger than $\Delta H^\circ$. The overall Gibbs Free Energy ($\Delta G^\circ$) plunges into the negative territory, and the reaction bursts into spontaneous action!

Enter Numerical Value:

Visualized Solution

$The Spontaneity Condition$

$Threshold Temperature$

$Calculating Δ H_{reaction}^{\circ}$

$Calculating Δ S_{reaction}^{\circ}$

$Unit Consistency & Substitution$

$Final Calculation$

$Thermodynamic Insight$

The Sigma Insight: Entropy and Free Energy

Decoding the Gibbs Free Energy Equation

Step 1

Step 2

The Crucial Trap

Finding the Threshold Temperature

The Physical Significance of the Result

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Spontaneity Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Threshold Temperature

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔHreaction∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔSreaction∘​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Unit Consistency & Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thermodynamic Insight