At what temperature, a reaction will be at equilibrium at 1 atm if `DeltaH and DeltaS` are 30.58 kJ and 66.1 J`k^-1` respectively? These values do not change with temperature in any significant fashion.

Options

A. 420.2 K
B. 462.6 K
C. 429 K
D. 642 K

Answer

To determine the temperature at which the reaction will be at equilibrium at 1 atm, we can use the relationship between Gibbs free energy (ΔG), enthalpy (ΔH), and entropy (ΔS). The condition for equilibrium is that ΔG = 0.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- ΔH = 30.58 kJ
- ΔS = 66.1 J/K

2. **Convert ΔH to Joules:**
Since ΔH is given in kilojoules, we need to convert it to joules for consistency with ΔS.
\[ \Delta H = 30.58 \, \text{kJ} \times 1000 \, \text{J/kJ} = 30580 \, \text{J} \]

3. **Set Up the Gibbs Free Energy Equation:**
The Gibbs free energy equation is given by:
\[ \Delta G = \Delta H - T \Delta S \]
For equilibrium, we set ΔG = 0:
\[ 0 = \Delta H - T \Delta S \]

4. **Rearrange the Equation to Solve for Temperature (T):**
\[ T \Delta S = \Delta H \]
\[ T = \frac{\Delta H}{\Delta S} \]

5. **Substitute the Values:**
\[ T = \frac{30580 \, \text{J}}{66.1 \, \text{J/K}} \]

6. **Calculate T:**
\[ T = 462.63 \, \text{K} \]

7. **Final Answer:**
The temperature at which the reaction will be at equilibrium at 1 atm is approximately:
\[ T \approx 462.6 \, \text{K} \]