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_fH^0(SnSO_2(s))=-5.81"KJ mol"^(-1), Delta_fH^0(CO_2(g))=-394.0"kJ mol"^(-1)`,
`S^0(SnO_2(s))=56.0JK^(-1)" mol"^(-1), S^0(Sn(s))=52.0J K^(-1)mol^(-1)`
`S^0(C(s))=6.0JK^(-1)mol^(-1), S^0(CO_2(g))=210.0JK^(-1)mol^(-1)`
Assume that the enthalpies and the entropies are temperature independent.

Options

A. 830
B. 865
C. 900
D. 935

Explanation

To determine the minimum temperature at which the reduction of cassiterite (SnO2) by coke (C) occurs, we will follow these steps:

### Step 1: Write the Reaction
The reduction of cassiterite by coke can be represented by the following chemical equation:
\[ \text{SnO}_2(s) + \text{C}(s) \rightarrow \text{Sn}(s) + \text{CO}_2(g) \]

### Step 2: Calculate the Enthalpy Change (ΔH)
Using the given standard enthalpies of formation:
- \( \Delta_fH^0(\text{SnO}_2(s)) = -581 \, \text{kJ/mol} \)
- \( \Delta_fH^0(\text{CO}_2(g)) = -394 \, \text{kJ/mol} \)
- The enthalpy of formation for elements in their standard state (Sn and C) is 0.

The enthalpy change for the reaction can be calculated as:
\[ \Delta H = \Delta_fH^0(\text{Sn}) + \Delta_fH^0(\text{CO}_2) - \Delta_fH^0(\text{SnO}_2) - \Delta_fH^0(\text{C}) \]
\[ \Delta H = 0 + (-394) - (-581) - 0 = -394 + 581 = 187 \, \text{kJ/mol} \]

### Step 3: Calculate the Entropy Change (ΔS)
Using the given standard entropies:
- \( S^0(\text{SnO}_2(s)) = 56.0 \, \text{J/K mol} \)
- \( S^0(\text{Sn}(s)) = 52.0 \, \text{J/K mol} \)
- \( S^0(\text{C}(s)) = 6.0 \, \text{J/K mol} \)
- \( S^0(\text{CO}_2(g)) = 210.0 \, \text{J/K mol} \)

The entropy change for the reaction can be calculated as:
\[ \Delta S = S^0(\text{Sn}) + S^0(\text{CO}_2) - S^0(\text{SnO}_2) - S^0(\text{C}) \]
\[ \Delta S = 52 + 210 - 56 - 6 = 210 \, \text{J/K mol} \]

### Step 4: Convert ΔH to J/mol
Since ΔS is in J/K mol, we need to convert ΔH from kJ to J:
\[ \Delta H = 187 \, \text{kJ/mol} = 187000 \, \text{J/mol} \]

### Step 5: Use the Gibbs Free Energy Equation
For a reaction to be spontaneous, the Gibbs free energy change (ΔG) must be less than 0:
\[ \Delta G = \Delta H - T \Delta S < 0 \]
Setting ΔG = 0 for the minimum temperature:
\[ 0 = \Delta H - T \Delta S \]
\[ T = \frac{\Delta H}{\Delta S} \]

### Step 6: Substitute Values
Substituting the values we have:
\[ T = \frac{187000 \, \text{J/mol}}{210 \, \text{J/K mol}} \approx 890.48 \, \text{K} \]

### Step 7: Conclusion
The minimum temperature at which the reduction of cassiterite by coke would take place is approximately:
\[ T \approx 890.48 \, \text{K} \]