Given the following data, calculate [tex]$\Delta G$[/tex] for the reaction below and state whether it is spontaneous or not:

[tex]\[CH_3CH_2OH + O_2 \rightarrow CH_3COOH + H_2O\][/tex]

[tex]\[
\begin{array}{l}
\Delta H = -495.2 \, \text{kJ} \\
\Delta S = -139 \, \text{J/K} \\
T = 298 \, \text{K} \\
\Delta G = \Delta H - T \Delta S
\end{array}
\][/tex]



Answer :

To solve the given problem, we will calculate the Gibbs free energy change ([tex]\(\Delta G\)[/tex]) using the provided values for enthalpy change ([tex]\(\Delta H\)[/tex]), entropy change ([tex]\(\Delta S\)[/tex]), and temperature ([tex]\(T\)[/tex]).

We are given the following data:
[tex]\[ \Delta H = -495.2 \ \text{kJ} \\ \Delta S = -139 \ \text{J/K} \\ T = 298 \ \text{K} \][/tex]

We need to use the formula:
[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]

### Step-by-Step Solution

1. Convert [tex]\(\Delta S\)[/tex] from J/K to kJ/K:

Recall that [tex]\(1 \ \text{kJ} = 1000 \ \text{J}\)[/tex]. Hence, to convert [tex]\(\Delta S\)[/tex] to kJ/K, we divide by 1000:
[tex]\[ \Delta S = -139 \ \text{J/K} \times \frac{1 \ \text{kJ}}{1000 \ \text{J}} = -0.139 \ \text{kJ/K} \][/tex]

2. Calculate [tex]\(\Delta G\)[/tex]:

Using the formula [tex]\(\Delta G = \Delta H - T \Delta S\)[/tex]:
[tex]\[ \Delta G = -495.2 \ \text{kJ} - (298 \ \text{K} \times -0.139 \ \text{kJ/K}) \][/tex]

3. Simplify the expression:

First, calculate the product [tex]\(T \Delta S\)[/tex]:
[tex]\[ T \Delta S = 298 \ \text{K} \times -0.139 \ \text{kJ/K} = -41.422 \ \text{kJ} \][/tex]

Now, substitute this back into the expression for [tex]\(\Delta G\)[/tex]:
[tex]\[ \Delta G = -495.2 \ \text{kJ} - (-41.422 \ \text{kJ}) = -495.2 \ \text{kJ} + 41.422 \ \text{kJ} = -453.778 \ \text{kJ} \][/tex]

4. Determine spontaneity:

A reaction is spontaneous if [tex]\(\Delta G\)[/tex] is less than 0. In this case:
[tex]\[ \Delta G = -453.778 \ \text{kJ} \][/tex]

Since [tex]\(\Delta G\)[/tex] is negative, the reaction is spontaneous.

### Conclusion:
- [tex]\(\Delta S\)[/tex] after conversion: [tex]\(-0.139 \ \text{kJ/K}\)[/tex]
- [tex]\(\Delta G\)[/tex]: [tex]\(-453.778 \ \text{kJ}\)[/tex]
- The reaction is spontaneous.