Q.At 450K, Kp= 2.0 × 10¹⁰/bar for the given reaction at equilibrium. 2SO2(g) + O2(g) ⇌ 2SO3

The key idea is converting between $K_p$ and $K_c$ using the relation $K_p = K_c (RT)^{\Delta n_g}$.

Step 1: Determine $\Delta n_g$, the change in moles of gas.

Step 2: Use the formula $K_p = K_c (RT)^{\Delta n_g}$. Here $R = 0.0831 \, \text{L bar mol}^{-1} \text{K}^{-1}$ and $T = 450 \, \text{K}$.

Step 3: Substitute the values. …

The equilibrium constant can be expressed in terms of partial pressures ($K_p$) or concentrations ($K_c$). The two are linked by the ideal gas law: for a gas-phase reaction, $K_p = K_c (RT)^{\Delta n_g}$, where $\Delta n_g$ is the difference between the sum of stoichiometric coefficients of gaseous products and reactants. This relation works because concentration $c = n/V = P/RT$, so each partial pressure term in $K_p$ can be replaced by $cRT$.

For the reaction $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$, the change in moles of gas is:

Since $\Delta n_g$ is negative, $K_c$ will be larger than $K_p$ by a factor of $RT$. Let’s work through the calculation step by step.

1. Write the conversion formula. The standard relation is:

Here, $R$ is the gas constant in appropriate units, $T$ is the temperature in Kelvin, and $\Delta n_g = -1$.

2. Rearrange to solve for $K_c$. Since $\Delta n_g = -1$, we have:

3. Plug in the values. Given $K_p = 2.0 \times 10^{10} \, \text{bar}^{-1}$ (note: the unit “per bar” comes from $\Delta n_g = -1$), $T = 450 \, \text{K}$, and $R = 0.08314 \, \text{L·bar·mol}^{-1}\text{K}^{-1}$ (the value that matches pressure in bar and volume in litres). Compute $RT$: $RT = 0.08314 \times 450 = 37.413 \, \text{L·bar·mol}^{-1}.$ …