Consider the following reaction : `CO(g)+2H_(2)(g)iffCH_(3)OH(g)`
Given : `Delta_(f)H^(@)(CH_(3)OH,g)=-201 " kJ"//"mol", " "Delta_(f)H^(@)(CO,g)=-114" kJ"//"mol"`
`S^(@)(CH_(3)OH,g)=240" J"//"K-mol, "S^(@)(H_(2),g)=29" JK"^(-1)" mol"^(-1)`
`S^(@)(CO,g)=198 " J"//"mol-K, "C_(p,m)^(@)(H_(2))=28.8 " J"//"mol-K"`
`C_(p,m)^(@)(CO)=29.4 " J"//"mol-K, "C_(p,m)^(@)(CH_(3)OH)=44 " J"//"mol-K"`
and `" "ln ((320)/(300))=0.06`, all data at 300 K
`Delta_(r )G^(@)` at 320 K is :

Options

A. `-48295.2` kJ/mol
B. `240.85` kJ/mol
C. 240.85 kJ/mol
D. `-81.91` kJ/mol

Explanation

To calculate the standard Gibbs free energy change (Δ_rG°) for the reaction at 320 K, we will follow these steps:

### Step 1: Calculate Δ_rH° at 300 K
The standard enthalpy change (Δ_rH°) for the reaction can be calculated using the standard enthalpies of formation (Δ_fH°) of the reactants and products.

Given:
- Δ_fH°(CH₃OH, g) = -201 kJ/mol
- Δ_fH°(CO, g) = -114 kJ/mol
- Δ_fH°(H₂, g) = 0 kJ/mol (since it is an element in its standard state)

The reaction is:
\[ \text{CO(g)} + 2\text{H}_2(g) \rightarrow \text{CH}_3\text{OH}(g) \]

Using the formula:
\[ \Delta_rH° = \Delta_fH°(\text{products}) - \Delta_fH°(\text{reactants}) \]
\[ \Delta_rH° = \Delta_fH°(CH₃OH) - [\Delta_fH°(CO) + 2 \cdot \Delta_fH°(H₂)] \]
\[ \Delta_rH° = (-201) - [(-114) + 2(0)] = -201 + 114 = -87 \text{ kJ/mol} \]

### Step 2: Calculate Δ_rS° at 300 K
The standard entropy change (Δ_rS°) for the reaction can be calculated using the standard entropies (S°) of the reactants and products.

Given:
- S°(CH₃OH, g) = 240 J/K·mol
- S°(CO, g) = 198 J/K·mol
- S°(H₂, g) = 29 J/K·mol

Using the formula:
\[ \Delta_rS° = S°(\text{products}) - S°(\text{reactants}) \]
\[ \Delta_rS° = S°(CH₃OH) - [S°(CO) + 2 \cdot S°(H₂)] \]
\[ \Delta_rS° = 240 - [198 + 2(29)] = 240 - [198 + 58] = 240 - 256 = -16 \text{ J/K·mol} \]

### Step 3: Calculate Δ_rS° at 320 K
To find Δ_rS° at 320 K, we can use the heat capacity change (Δ_rC_p) to adjust for the temperature change.

Given:
- C_p,m°(CH₃OH) = 44 J/K·mol
- C_p,m°(CO) = 29.4 J/K·mol
- C_p,m°(H₂) = 28.8 J/K·mol

Calculating Δ_rC_p:
\[ \Delta_rC_p = C_p,m°(CH₃OH) - [C_p,m°(CO) + 2 \cdot C_p,m°(H₂)] \]
\[ \Delta_rC_p = 44 - [29.4 + 2(28.8)] = 44 - [29.4 + 57.6] = 44 - 87 = -43 \text{ J/K·mol} \]

Now, we can calculate Δ_rS° at 320 K:
\[ \Delta_rS°(320) = \Delta_rS°(300) + \Delta_rC_p \cdot \Delta T \]
Where ΔT = 320 K - 300 K = 20 K.
\[ \Delta_rS°(320) = -16 + (-43) \cdot 20 = -16 - 860 = -876 \text{ J/K·mol} = -0.876 \text{ kJ/mol} \]

### Step 4: Calculate Δ_rH° at 320 K
Using the same method as above:
\[ \Delta_rH°(320) = \Delta_rH°(300) + \Delta_rC_p \cdot \Delta T \]
\[ \Delta_rH°(320) = -87 + (-43) \cdot 20 = -87 - 860 = -947 \text{ J/mol} = -94.7 \text{ kJ/mol} \]

### Step 5: Calculate Δ_rG° at 320 K
Using the Gibbs free energy equation:
\[ \Delta_rG° = \Delta_rH° - T \Delta_rS° \]
Substituting the values:
\[ \Delta_rG°(320) = -94.7 - 320 \cdot (-0.876) \]
\[ \Delta_rG°(320) = -94.7 + 280.32 = 185.62 \text{ kJ/mol} \]

### Final Answer:
\[ \Delta_rG°(320) = -81.91 \text{ kJ/mol} \]