A reaction has AH°= 100 kJ/mol and AS°= 250 J/mol.K. Is the reaction spontaneous at room temperature? If not, unde what temperature conditions will it become spontaneous?

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### Spontaneity of a Chemical Reaction

A chemical reaction has the following thermodynamic parameters:
- **ΔH° (Enthalpy change)** = 100 kJ/mol
- **ΔS° (Entropy change)** = 250 J/mol·K

The problem posed is: 

- **Question**: Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?

### Explanation

To determine if a reaction is spontaneous, we use the Gibbs free energy equation:

\[ \Delta G° = \Delta H° - T \Delta S° \]

Where:
- \(\Delta G°\) is the change in Gibbs free energy.
- \(\Delta H°\) is the change in enthalpy.
- \(T\) is the temperature in Kelvin.
- \(\Delta S°\) is the change in entropy.

A reaction is spontaneous if \(\Delta G° < 0\).

### Step-by-Step Solution

1. **Convert Units:** 
   Make sure that the units for ΔH° and ΔS° are compatible. ΔS° is given in J/mol·K, while ΔH° is in kJ/mol. Convert ΔH° to J/mol:

   \[ \Delta H° = 100 \text{ kJ/mol} \times 1000 \text{ J/kJ} = 100,000 \text{ J/mol} \]

2. **Set Up Gibbs Free Energy Equation:**
   Substitute the values into the equation \(\Delta G° = \Delta H° - T \Delta S°\):

   \[ \Delta G° = 100,000 \text{ J/mol} - T \times 250 \text{ J/mol·K} \]

3. **Determine Spontaneity at Room Temperature:**
   Assume room temperature is approximately 298 K. Substitute T = 298 K into the equation:

   \[ \Delta G° = 100,000 \text{ J/mol} - 298 \text{ K} \times 250 \text{ J/mol·K} \]
   \[ \Delta G° = 100,000 \text{ J/mol} - 74,500 \text{ J/mol} \]
   \[ \Delta G° = 25,500 \text{ J/mol} \]

   Since \(\Delta G° >
Transcribed Image Text:### Spontaneity of a Chemical Reaction A chemical reaction has the following thermodynamic parameters: - **ΔH° (Enthalpy change)** = 100 kJ/mol - **ΔS° (Entropy change)** = 250 J/mol·K The problem posed is: - **Question**: Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous? ### Explanation To determine if a reaction is spontaneous, we use the Gibbs free energy equation: \[ \Delta G° = \Delta H° - T \Delta S° \] Where: - \(\Delta G°\) is the change in Gibbs free energy. - \(\Delta H°\) is the change in enthalpy. - \(T\) is the temperature in Kelvin. - \(\Delta S°\) is the change in entropy. A reaction is spontaneous if \(\Delta G° < 0\). ### Step-by-Step Solution 1. **Convert Units:** Make sure that the units for ΔH° and ΔS° are compatible. ΔS° is given in J/mol·K, while ΔH° is in kJ/mol. Convert ΔH° to J/mol: \[ \Delta H° = 100 \text{ kJ/mol} \times 1000 \text{ J/kJ} = 100,000 \text{ J/mol} \] 2. **Set Up Gibbs Free Energy Equation:** Substitute the values into the equation \(\Delta G° = \Delta H° - T \Delta S°\): \[ \Delta G° = 100,000 \text{ J/mol} - T \times 250 \text{ J/mol·K} \] 3. **Determine Spontaneity at Room Temperature:** Assume room temperature is approximately 298 K. Substitute T = 298 K into the equation: \[ \Delta G° = 100,000 \text{ J/mol} - 298 \text{ K} \times 250 \text{ J/mol·K} \] \[ \Delta G° = 100,000 \text{ J/mol} - 74,500 \text{ J/mol} \] \[ \Delta G° = 25,500 \text{ J/mol} \] Since \(\Delta G° >
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