Question

Equilibrium involving SO2(g), O2(g) & SO3(g) is important in sulfuric acid production. When a 0.0200 mol sample of SO3 is introduced into an evacuated 1.52 L vessel at 900 K, 0.0142 mol, SO3 is found to be present at equilibrium. calculate the value of Kp for the dissociation of SO3(g) at 900 K?

Answer 1

Answer on Question #73501 – Chemistry – Inorganic Chemistry

Equilibrium involving $\mathrm{SO}_2(\mathrm{g})$, $\mathrm{O}_2(\mathrm{g})$ and $\mathrm{SO}_3(\mathrm{g})$ is important in sulfuric acid production. When a 0.0200 mol sample of $\mathrm{SO}_3$ is introduced into an evacuated $1.52\mathrm{L}$ vessel at $900\mathrm{K}$, 0.0142 mol, $\mathrm{SO}_3$ is found to be present at equilibrium. Calculate the value of $K_{\mathrm{p}}$ for the dissociation of $\mathrm{SO}_3(\mathrm{g})$ at $900\mathrm{K}$?

Solution:

$$2 \mathrm{SO}_3(\mathrm{g}) \rightarrow 2 \mathrm{SO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g})$$

1. Calculate the amount of $\mathrm{SO}_3$ which has reacted away:

$\Delta n(\mathrm{SO}_3) = n(\mathrm{SO}_3)_{\text{initial}} - n(\mathrm{SO}_3)_{\text{equilibrium}} = 0.0200 \, \mathrm{mol} - 0.0142 \, \mathrm{mol} = 0.0058 \, \mathrm{mol}$

2. Calculate the amount of sulfur dioxide and oxygen.

Dissociation reaction is given by:

$$\mathrm{SO}_3(\mathrm{g}) \leftarrow \rightarrow \mathrm{SO}_2(\mathrm{g}) + \left(\frac{1}{2}\right) \mathrm{O}_2(\mathrm{g})$$

So one mole of $\mathrm{SO}_2$ and one half mole of $\mathrm{O}_2$ is formed per mole of $\mathrm{SO}_3$ reacted away.

Hence:

$$n(\mathrm{SO}_2)_{\text{equilibrium}} = \Delta n(\mathrm{SO}_3) = 0.0058 \, \mathrm{mol}$$ $$n(\mathrm{O}_2)_{\text{equilibrium}} = \left(\frac{1}{2}\right) \cdot \Delta n(\mathrm{SO}_3) = 0.0029 \, \mathrm{mol}$$

3. Calculate equilibrium partial pressures.

Assuming ideal gas mixture partial pressures are given by

$$p(i) = n(i) \cdot R \cdot T / V$$

(R - universal gas constant, T - absolute temperature, V - Volume)

$$p(\mathrm{SO}_3) = 0.082 \, \mathrm{L} \cdot \mathrm{atm} / (K \cdot \mathrm{mol}) \times 900 \, K \times 0.0142 \, \mathrm{mol} / 1.52 \, \mathrm{L} = 48.6 \, \mathrm{atm/mol} \times 0.0142 \, \mathrm{mol} = 0.690 \, \mathrm{atm}$$ $$p(\mathrm{SO}_2) = 48.6 \, \mathrm{atm/mol} \times 0.0058 \, \mathrm{mol} = 0.282 \, \mathrm{atm}$$ $$p(\mathrm{O}_2) = 48.6 \, \mathrm{atm/mol} \times 0.0029 \, \mathrm{mol} = 0.141 \, \mathrm{atm}$$

4. Calculate equilibrium constant

According to reaction equation:

$$K_p = p^2(\mathrm{SO}_2) \times p(\mathrm{O}_2) / p^2(\mathrm{SO}_3) = (0.282)^2 \times 0.141 / (0.690)^2 = 0.0235 = 0.024$$

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