Problem 28 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in \(\mathrm{K}\) ) at which the reduction of cassiterite by coke would take place. [Adv. 2020] At \(298 \mathrm{~K}: \Delta_{f} H^{0}\left(\mathrm{SnO}_{2}(s)\right)=-581.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\), \(\Delta_{f} H^{0}\left(\mathrm{CO}_{2}(g)\right)=-394.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(S^{0}\left(\mathrm{SnO}_{2}(s)\right)=56.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\), \(S^{0}(\operatorname{Sn}(s))=52.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) \(S^{0}(\mathrm{C}(s))=6.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \quad S^{0}\left(\mathrm{CO}_{2}(g)\right)=210.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\). Assume that the enthalpies and the entropies are temperature independent.

Step-by-Step Solution

Answer

The minimum temperature for the reduction is approximately 512.4 K.

Step 1: Write Down the Reaction Equation

The reaction for the reduction of cassiterite (SnO2) with coke (C) to form tin (Sn) and carbon dioxide (CO2) is: \[ \text{SnO}_2(s) + 2 \text{C}(s) \rightarrow \text{Sn}(s) + 2 \text{CO}_2(g) \]

Step 2: Calculate Change in Standard Enthalpy (ΔH°)

Use the standard enthalpy of formation values to calculate the change in enthalpy for the reaction: \[ \Delta H^0 = [\Delta_f H^0(\text{Sn}(s)) + 2\Delta_f H^0(\text{CO}_2(g))] - [\Delta_f H^0(\text{SnO}_2(s)) + 2\Delta_f H^0(\text{C}(s))] \] Since the standard enthalpy of formation for elemental solids (Sn and C) is zero, this simplifies to: \[ \Delta H^0 = [0 + 2(-394.0)] - [-581.0 + 2(0)] = -788.0 + 581.0 = -207.0 \text{kJ/mol} \]

Step 3: Calculate Change in Standard Entropy (ΔS°)

Use the standard entropy values to find the change in entropy for the reaction: \[ \Delta S^0 = [S^0(\text{Sn}(s)) + 2S^0(\text{CO}_2(g))] - [S^0(\text{SnO}_2(s)) + 2S^0(\text{C}(s))] \] Calculating gives: \[ \Delta S^0 = [52.0 + 2(210.0)] - [56.0 + 2(6.0)] = 472.0 - 68.0 = 404.0 \text{J/K/mol} \]

Step 4: Find Minimum Temperature Using Gibbs Free Energy

Use the Gibbs free energy equation where keeping \(\Delta G^0 = 0\): \[ \Delta G^0 = \Delta H^0 - T\Delta S^0 \] Setting \(\Delta G^0 = 0\), you get: \[ 0 = -207.0 \times 10^3 - T(404.0) \] Solving for \(T\): \[ 207.0 \times 10^3 = T \times 404 \] \[ T = \frac{207.0 \times 10^3}{404} \approx 512.4 \text{K} \]

Step 5: Conclusion and Interpretation

The minimum temperature at which the reduction of cassiterite by coke can take place, based on the thermodynamic data provided, is approximately 512.4 K.

Key Concepts

Reduction Reactions

Reduction reactions are a key component of chemical processes where a substance gains electrons. In our case, tin is obtained by reducing cassiterite (SnO_2) with coke (carbon). This process is crucial in metallurgy where ores are converted into their metallic states.
In this specific reaction, coke serves as the reducing agent by supplying electrons, transforming SnO_2 to pure tin. Simultaneously, the C in coke is oxidized to CO_2 gas. This reaction can be expressed as:
\[ \text{SnO}_2(s) + 2 \text{C}(s) \rightarrow \text{Sn}(s) + 2 \text{CO}_2(g) \]Reduction reactions are critical because they allow the extraction of metals from naturally occurring minerals. Understanding which substances can act as reducing agents and how the reactions occur underpins the efficient design of industrial processes.

Enthalpy Calculation

Enthalpy (H) in thermodynamics refers to the total heat content of a system. It is an essential concept when analyzing reactions as it indicates whether a process is endothermic or exothermic.
In calculating enthalpy changes for this reaction, we use standard enthalpies of formation (\( \Delta_f H^{0} \)), which are the heat changes when one mole of a compound forms from its elements in their standard states. The formula we use is:
\[ \Delta H^0 = [\Delta_f H^0(\text{Sn}(s)) + 2\Delta_f H^0(\text{CO}_2(g))] - [\Delta_f H^0(\text{SnO}_2(s)) + 2\Delta_f H^0(\text{C}(s))] \]
This formula simplifies as the standard enthalpies for elements (Sn and C) are zero:

• \( \Delta_f H^{0} (SnO_2) = -581.0 \text{ kJ/mol} \)
• \( \Delta_f H^{0}(CO_2) = -394.0 \text{ kJ/mol} \)

This calculation shows the reaction is exothermic because \( \Delta H^{0} = -207.0 \text{ kJ/mol} \). This means energy is released during the reaction, aiding product formation.

Entropy Change

Entropy (S) is a measure of disorder or randomness in a system, and changes in entropy give important insights into the feasibility of a reaction.
To find the change in standard entropy (\( \Delta S^0 \)), we look at the standard entropies of reactants and products. The formula for \( \Delta S^0 \) in our reaction is:
\[ \Delta S^0 = [S^0(\text{Sn}(s)) + 2S^0(\text{CO}_2(g))] - [S^0(\text{SnO}_2(s)) + 2S^0(\text{C}(s))] \]
The individual entropy values used are:

• \( S^0(\text{SnO}_2) = 56.0 \text{ J/K/mol} \)
• \( S^0(Sn) = 52.0 \text{ J/K/mol} \)
• \( S^0(C) = 6.0 \text{ J/K/mol} \)
• \( S^0(\text{CO}_2) = 210.0 \text{ J/K/mol} \)

By calculating \( \Delta S^{0} = 404.0 \text{ J/K/mol} \), it shows an increase in disorder as the number of gas molecules increases from reactants to products, generally favoring spontaneous reactions.

Gibbs Free Energy

Gibbs Free Energy (G) combines enthalpy and entropy to predict the spontaneity of a reaction at constant temperature and pressure. It's represented by the equation:
\[ \Delta G^0 = \Delta H^0 - T\Delta S^0 \]
Calculating Gibbs Free Energy for our reaction helps us determine the minimum temperature for spontaneity, set by \( \Delta G^0 = 0 \). This gives:
\[ 0 = -207.0 \times 10^3 - T(404.0) \]
By rearranging and solving for \( T \):
\[ T = \frac{207.0 \times 10^3}{404} \approx 512.4 \text{ K} \]
This temperature represents the balance point where both enthalpy and entropy favor the reduction reaction. At temperatures above this, the reaction is more likely to proceed spontaneously, verifying the importance of both energy content and randomness in determining reaction direction.