Question

For the reaction, $ SO_2 ( g) + \frac{1}{2} O_2 \, ( g) \rightleftharpoons SO_3 \, ( g) $ if $ K_P = K_C \, ( RT)^x $ X where, the symbols have usual meaning, then the value of x is (assuming ideality)

• A. -1
• B. - $ \frac{1}{2} $
• C. $ \frac{1}{2} $
• D. 1

Answer 1

The Correct Option is B

 To solve the problem, we need to find the relationship between \(K_P\) (the equilibrium constant calculated with partial pressures) and \(K_C\) (the equilibrium constant calculated with concentrations) for the given reaction:

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

The general relation between \(K_P\) and \(K_C\) for a gaseous reaction is given by:

\(K_P = K_C (RT)^{\Delta n}\)

Here, \(\Delta n\) is the change in the number of moles of gas during the reaction, that is the difference between the total number of moles of gaseous products and the total number of moles of gaseous reactants.

Step-by-Step Calculation:

1. Identify the number of moles of gaseous products and reactants.
   • Products: \(1\) mole of \(SO_3\)
   • Reactants: \(1\) mole of \(SO_2\) + \(\frac{1}{2}\) mole of \(O_2\)
   • Total moles of reactants = \(1 + \frac{1}{2} = \frac{3}{2}\)
2. Calculate \(\Delta n\):
   • \(\Delta n = \text{moles of products} - \text{moles of reactants} = 1 - \frac{3}{2} = -\frac{1}{2}\)
3. Substitute \(\Delta n\) into the formula:
   • \(K_P = K_C (RT)^{- \frac{1}{2}}\)

From the equation, we observe that \(x = -\frac{1}{2}\). Hence, the value of \(x\) is \(-\frac{1}{2}\), which matches the correct option:

- \(-\frac{1}{2}\)

.

 

Thus, the correct answer is \(-\frac{1}{2}\).