Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the m

Q: Why do elements have non-zero standard entropy values if their enthalpy of formation is zero?

Entropy is an absolute quantity based on the Third Law of Thermodynamics, which states that a perfect crystal at 0 K has zero entropy. At 298 K, all substances, including elements, have some thermal energy and thus a positive absolute entropy.

Q: Why did we multiply the enthalpy change by 1000?

The standard enthalpy of formation is given in kilojoules per mole (kJ/mol), whereas standard entropy is given in joules per kelvin per mole (J/K·mol). To use them together in the Gibbs free energy equation, they must be in the exact same energy units.

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

An entropy-driven reaction is one where the enthalpy change is positive (endothermic, unfavorable) but the entropy change is also positive (favorable). At high temperatures, the $T\Delta S$ term dominates, making $\Delta G$ negative and driving the reaction forward.

Q: Why is the condition for minimum temperature $T > \Delta H / \Delta S$ and not $T = \Delta H / \Delta S$?

At exactly $T = \Delta H / \Delta S$, the system is in equilibrium ($\Delta G = 0$). For the forward reaction to be spontaneous and actually proceed, $\Delta G$ must be strictly less than zero, which requires the temperature to be slightly greater than this equilibrium value.

Question

Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place.
At 298 K : $\Delta_f H^\circ(	ext{SnO}_2(s)) = -581.0 	ext{ kJ mol}^{-1}$, $\Delta_f H^\circ(	ext{CO}_2(g)) = -394.0 	ext{ kJ mol}^{-1}$
$S^\circ(	ext{SnO}_2(s)) = 56.0 	ext{ J K}^{-1} 	ext{ mol}^{-1}$, $S^\circ(	ext{Sn}(s)) = 52.0 	ext{ J K}^{-1} 	ext{ mol}^{-1}$,
$S^\circ(	ext{C}(s)) = 6.0 	ext{ J K}^{-1} 	ext{ mol}^{-1}$, $S^\circ(	ext{CO}_2(g)) = 210.0 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.
Assume that the enthalpies and the entropies are temperature independent.

Sigma Percentile · Accepted Answer

### The Fiery Crucible: Unlocking Tin with Thermodynamics

The extraction of metals from their ores is one of the oldest applications of chemistry in human history. But behind the roaring furnaces and glowing molten metal lies a beautiful, invisible dance of thermodynamic properties. In this problem, we are tasked with finding the minimum temperature required to extract tin ($	ext{Sn}$) from its ore, cassiterite ($	ext{SnO}_2$), using coke (carbon) as a reducing agent.

To solve this, we must consult the ultimate arbiter of chemical destiny: the Gibbs Free Energy equation.

### Setting the Stage: The Chemical Reaction

Before we can crunch any numbers, we need to know exactly what is happening inside the furnace. We are reacting solid tin dioxide with solid carbon to produce solid tin and gaseous carbon dioxide. The balanced chemical equation is:

$$ 	ext{SnO}_2(s) + 	ext{C}(s) \longrightarrow 	ext{Sn}(s) + 	ext{CO}_2(g) $$

Notice something critical here: we are starting with two solids and producing a solid and a gas. This phase change from solid to gas is a massive hint about the role entropy will play in this reaction.

### The Energy Barrier: Enthalpy ($\Delta H^\circ$)

First, let's determine the heat energy required for this reaction, known as the standard enthalpy of reaction ($\Delta H^\circ_{	ext{rxn}}$). We calculate this by taking the sum of the enthalpies of formation of the products and subtracting the sum of the enthalpies of formation of the reactants.

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

We are given the formation enthalpies for $	ext{CO}_2$ and $	ext{SnO}_2$. But what about elemental tin and carbon? By thermodynamic convention, the standard enthalpy of formation for any element in its most stable reference state is exactly zero.

Substituting the given values:

$$ \Delta H^\circ_{	ext{rxn}} = [-394.0] - [-581.0] = +187.0 	ext{ kJ mol}^{-1} $$

The positive sign is crucial. It tells us the reaction is **endothermic**. It requires an input of $187.0 	ext{ kJ}$ of energy per mole to proceed. From an enthalpy perspective, this reaction does not want to happen on its own.

### The Agent of Chaos: Entropy ($\Delta S^\circ$)

If enthalpy is fighting against us, what is driving the reaction forward? Enter entropy ($\Delta S^\circ$), the measure of randomness or disorder in the system. We calculate the standard entropy of reaction similarly:

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

Unlike enthalpy, elements do have absolute entropy values at $298 	ext{ K}$ because they possess thermal energy (Third Law of Thermodynamics). Plugging in the values:

$$ \Delta S^\circ_{	ext{rxn}} = [52.0 + 210.0] - [56.0 + 6.0] $$

$$ \Delta S^\circ_{	ext{rxn}} = 262.0 - 62.0 = +200.0 	ext{ J K}^{-1} 	ext{ mol}^{-1} $$

The entropy change is highly positive! This makes perfect physical sense because we are generating a gas ($	ext{CO}_2$) from solid reactants. Gases have significantly more microstates and randomness than solids.

### The Master Equation: Gibbs Free Energy

We have a conflict: enthalpy opposes the reaction, but entropy favors it. Who wins? The Gibbs Free Energy ($\Delta G^\circ$) decides. For a reaction to be spontaneous, $\Delta G^\circ$ must be strictly less than zero.

$$ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ < 0 $$

Rearranging this inequality to solve for temperature ($T$):

$$ T > \frac{\Delta H^\circ}{\Delta S^\circ} $$

This tells us that at high temperatures, the $T\Delta S^\circ$ term will eventually overpower the positive $\Delta H^\circ$, making the overall $\Delta G^\circ$ negative. This is the hallmark of an **entropy-driven** reaction.

### The Final Calculation: Beware the Units

Now, we substitute our calculated values to find the minimum temperature. But there is a classic trap here! The enthalpy is given in kilojoules ($	ext{kJ}$), while the entropy is in joules ($	ext{J}$). We must convert the enthalpy to joules by multiplying by $1000$.

$$ T > \frac{187.0 	imes 1000 	ext{ J mol}^{-1}}{200.0 	ext{ J K}^{-1} 	ext{ mol}^{-1}} $$

$$ T > \frac{187000}{200} $$

$$ T > 935 	ext{ K} $$

Therefore, the furnace must be heated to a minimum of **$935 	ext{ K}$** for the reduction of cassiterite by coke to become spontaneous. Thermodynamics beautifully dictates the exact conditions required to unlock the metal from its ore.

Enter Numerical Value:

Visualized Solution

\text{The Chemical Reaction}

\text{Enthalpy of Reaction } (\Delta H^\circ_{\text{rxn}})

\text{Calculating } \Delta H^\circ_{\text{rxn}}

\text{Entropy of Reaction } (\Delta S^\circ_{\text{rxn}})

\text{Calculating } \Delta S^\circ_{\text{rxn}}

\text{Gibbs Free Energy \& Spontaneity}

\text{Calculating Minimum Temperature}

\text{Conclusion}

The Sigma Insight: Entropy and Free Energy

The Fiery Crucible

Setting the Stage

The Energy Barrier

The Agent of Chaos

The Master Equation

The Final Calculation

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