Given below are the values of `DeltaH^(ɵ)` and `DeltaS^(ɵ)` for the reaction given below at `27^(@)C`.
`SO_(2)(g)+1/2O_(2)(g) rarr SO_(3)(g)`
`DeltaH^(ɵ)=-98.32 kJ mol^(-1), DeltaS^(ɵ)=-95 J mol^(-1)`
Find `K_(p)` for the reaction

Explanation

To find the equilibrium constant \( K_p \) for the reaction

\[ SO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons SO_3(g) \]

given the values of \( \Delta H^{\circ} \) and \( \Delta S^{\circ} \) at \( 27^{\circ}C \), we can follow these steps:

### Step 1: Convert Temperature to Kelvin
The temperature in Celsius is given as \( 27^{\circ}C \). To convert this to Kelvin:

\[ T(K) = 27 + 273.15 = 300.15 \approx 300 \text{ K} \]

### Step 2: Convert \( \Delta H^{\circ} \) to Joules
Given \( \Delta H^{\circ} = -98.32 \, \text{kJ/mol} \), we convert this to Joules:

\[ \Delta H^{\circ} = -98.32 \times 1000 = -98320 \, \text{J/mol} \]

### Step 3: Use the Gibbs Free Energy Equation
The Gibbs free energy change at standard conditions is given by:

\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]

We know \( \Delta S^{\circ} = -95 \, \text{J/mol} \). Now, substituting the values:

\[ \Delta G^{\circ} = -98320 \, \text{J/mol} - (300 \, \text{K})(-95 \, \text{J/mol}) \]

Calculating \( T \Delta S^{\circ} \):

\[ T \Delta S^{\circ} = 300 \times -95 = -28500 \, \text{J/mol} \]

Now substituting back:

\[ \Delta G^{\circ} = -98320 + 28500 = -69820 \, \text{J/mol} \]

### Step 4: Relate \( \Delta G^{\circ} \) to \( K_p \)
Using the relationship:

\[ \Delta G^{\circ} = -RT \ln K_p \]

Where \( R = 8.314 \, \text{J/(mol K)} \). Substituting the values we have:

\[ -69820 = -8.314 \times 300 \ln K_p \]

### Step 5: Solve for \( \ln K_p \)
Rearranging gives:

\[ \ln K_p = \frac{69820}{8.314 \times 300} \]

Calculating the denominator:

\[ 8.314 \times 300 = 2494.2 \]

Now substituting:

\[ \ln K_p = \frac{69820}{2494.2} \approx 27.99 \]

### Step 6: Calculate \( K_p \)
Taking the antilogarithm:

\[ K_p = e^{27.99} \approx 1.4 \times 10^{12} \]

### Step 7: Determine the Units of \( K_p \)
To find the units of \( K_p \), we calculate \( \Delta n \):

\[ \Delta n = n_{products} - n_{reactants} = 1 - \left(1 + \frac{1}{2}\right) = 1 - 1.5 = -0.5 \]

Thus, the units of \( K_p \) are:

\[ K_p \text{ (units)} = \text{atm}^{\Delta n} = \text{atm}^{-0.5} \]

### Final Answer
The value of \( K_p \) for the reaction is:

\[ K_p \approx 1.4 \times 10^{12} \, \text{atm}^{-0.5} \]

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