For the reaction `2A_3 hArr 3A_2` the equilibrium constant and the `DeltaG^@` values at a certain temperature are respectively `1 xx 10^30` and -172.4 kJ `"mol"^(-1)` . The equilibrium temperature in `.^@C` is about (2.303 R =19.15 J `K^(-1) mol^(-1)`)

Options

A. 300
B. 27
C. 273
D. 298

Answer

To find the equilibrium temperature for the reaction \( 2A_3 \rightleftharpoons 3A_2 \) given the equilibrium constant \( K = 1 \times 10^{30} \) and \( \Delta G^\circ = -172.4 \, \text{kJ mol}^{-1} \), we can use the relationship between the Gibbs free energy change and the equilibrium constant, which is given by the equation:

\[ \Delta G^\circ = -RT \ln K \]

Where:
- \( \Delta G^\circ \) is the standard Gibbs free energy change,
- \( R \) is the universal gas constant (in J K\(^{-1}\) mol\(^{-1}\)),
- \( T \) is the temperature in Kelvin,
- \( K \) is the equilibrium constant.

### Step 1: Convert \( \Delta G^\circ \) to J/mol
Since \( R \) is given in J K\(^{-1}\) mol\(^{-1}\), we need to convert \( \Delta G^\circ \) from kJ to J:

\[ \Delta G^\circ = -172.4 \, \text{kJ mol}^{-1} \times 1000 \, \text{J kJ}^{-1} = -172400 \, \text{J mol}^{-1} \]

### Step 2: Substitute values into the equation
Now we can substitute the values into the equation \( \Delta G^\circ = -RT \ln K \):

\[ -172400 = - (19.15) T \ln(1 \times 10^{30}) \]

### Step 3: Calculate \( \ln K \)
Calculate \( \ln(1 \times 10^{30}) \):

\[ \ln(1 \times 10^{30}) = \ln(10^{30}) = 30 \ln(10) \approx 30 \times 2.303 = 69.09 \]

### Step 4: Substitute \( \ln K \) into the equation
Now substitute \( \ln K \) back into the equation:

\[ -172400 = - (19.15) T (69.09) \]

### Step 5: Solve for \( T \)
Rearranging the equation to solve for \( T \):

\[ 172400 = 19.15 \times T \times 69.09 \]

\[ T = \frac{172400}{19.15 \times 69.09} \]

Calculating the denominator:

\[ 19.15 \times 69.09 \approx 1326.36 \]

Now calculate \( T \):

\[ T \approx \frac{172400}{1326.36} \approx 129.94 \, \text{K} \]

### Step 6: Convert \( T \) to Celsius
To convert from Kelvin to Celsius:

\[ T(°C) = T(K) - 273.15 \]

\[ T(°C) \approx 129.94 - 273.15 \approx -143.21 \, °C \]

### Final Answer
The equilibrium temperature is approximately \( -143.21 \, °C \).