Standard Gibb's energy of reaction `(Delta_(r )G^(@))` at a certain temperature can be computed `Delta_(r )G^(@)=Delta_(r)H^(@)-T.Delta_(r )S^(@)` and the change in the value of `Delta_(r)H^(@)` and `Delta_(r)S^(@)` for a reaction with temperature can be computed as follows :
`Delta_(r )H_(T_(2))^(@)-Delta_(r )H_(T_(1))^(@)=Delta_(r )C_(p)^(@)(T_(2)-T_(1))`
`Delta_(r )S_(T_(2))^(@)-Delta_(r )S_(T_(1))^(@)=Delta_(r )C_(p)^(@)ln.(T_(2)/T_(1))`
`" "Delta_(r )G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@)`
and `" by "Delta_(r )G^(@)=-"RT " ln K_(eq)`.
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 )S^(@)` at 320 K is :

Options

A. 155.18 J/mol-K
B. 150.02 J/mol-K
C. 172 J/mol-K
D. None of these

Explanation

To calculate the change in standard entropy (\( \Delta_r S^\circ \)) at 320 K for the reaction \( \text{CO(g)} + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g) \), we will follow these steps:

### Step 1: Calculate \( \Delta_r S^\circ \) at 300 K

The standard entropy change for the reaction at 300 K can be calculated using the formula:

\[ \Delta_r S^\circ_{300} = S^\circ(\text{CH}_3\text{OH}) - S^\circ(\text{CO}) - 2 \cdot S^\circ(\text{H}_2) \]

Given:
- \( S^\circ(\text{CH}_3\text{OH}) = 240 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ(\text{CO}) = 198 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( S^\circ(\text{H}_2) = 29 \, \text{J K}^{-1} \text{mol}^{-1} \)

Substituting the values:

\[ \Delta_r S^\circ_{300} = 240 - 198 - 2 \cdot 29 \]
\[ \Delta_r S^\circ_{300} = 240 - 198 - 58 \]
\[ \Delta_r S^\circ_{300} = 240 - 256 = -16 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 2: Calculate \( \Delta_r C_p^\circ \)

Next, we calculate the change in heat capacity (\( \Delta_r C_p^\circ \)):

\[ \Delta_r C_p^\circ = C_{p,m}^\circ(\text{CH}_3\text{OH}) - C_{p,m}^\circ(\text{CO}) - 2 \cdot C_{p,m}^\circ(\text{H}_2) \]

Given:
- \( C_{p,m}^\circ(\text{CH}_3\text{OH}) = 44 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( C_{p,m}^\circ(\text{CO}) = 29.4 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( C_{p,m}^\circ(\text{H}_2) = 28.8 \, \text{J K}^{-1} \text{mol}^{-1} \)

Substituting the values:

\[ \Delta_r C_p^\circ = 44 - 29.4 - 2 \cdot 28.8 \]
\[ \Delta_r C_p^\circ = 44 - 29.4 - 57.6 \]
\[ \Delta_r C_p^\circ = 44 - 87 = -43 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Step 3: Calculate \( \Delta_r S^\circ \) at 320 K

Using the formula for the change in entropy with temperature:

\[ \Delta_r S^\circ_{320} - \Delta_r S^\circ_{300} = \Delta_r C_p^\circ \cdot \ln\left(\frac{T_2}{T_1}\right) \]

Where \( T_1 = 300 \, \text{K} \) and \( T_2 = 320 \, \text{K} \):

Substituting the values:

\[ \Delta_r S^\circ_{320} - (-16) = -43 \cdot \ln\left(\frac{320}{300}\right) \]

Given \( \ln\left(\frac{320}{300}\right) = 0.06 \):

\[ \Delta_r S^\circ_{320} + 16 = -43 \cdot 0.06 \]
\[ \Delta_r S^\circ_{320} + 16 = -2.58 \]
\[ \Delta_r S^\circ_{320} = -2.58 - 16 \]
\[ \Delta_r S^\circ_{320} = -18.58 \, \text{J K}^{-1} \text{mol}^{-1} \]

### Final Answer

Thus, the change in standard entropy \( \Delta_r S^\circ \) at 320 K is:

\[ \Delta_r S^\circ_{320} = -18.58 \, \text{J K}^{-1} \text{mol}^{-1} \]

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