**We are given the following reaction:**

**SO**_{2}_{(g)} + 2 H_{2}S_{(g)} ⇌ 3 S_{(s) }+ 2 H_{2}O_{(g)}

**We are asked to do the following:**

**Determine the equilibrium constant, k, for the reaction.****Calculate the equilibrium pressure of SO**_{2(g) }when Pressure of H_{2}S = Pressure of SO_{2}and the vapor pressure of water is 26 torr.

**Find the equilibrium constant, k.**

Recall that ** ΔG˚_{rxn} and K** are related to each other:

$\overline{){\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{RTlnK}}}$

We can use the following equation to solve for ** ΔG˚_{rxn}**:

$\overline{){\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{prod}}{\mathbf{-}}{\mathbf{\Delta G}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{react}}}$

**In determining k, we need to do the following steps:**

* Step 1:* Calculate for ΔG˚

* Step 2:* Calculate for K.

**Step 1***:* Calculate for **ΔG˚**_{rxn}:

**Given:**

ΔG˚_{f}, SO_{2}_{(g)} = **– 300.4 kJ/mol**

ΔG˚_{f}, H_{2}S_{(g)} = **– 33.01 kJ/mol**

ΔG˚_{f}, S_{(s)} = **0 kJ/mol**

ΔG˚_{f}, H_{2}O_{(g)} = **– 228.57 kJ/mol**

The reaction SO_{2}(g) + 2H_{2}S(g) ⇌ 3S(s) + 2H_{2}O(g) is the basis of a suggested method for removal of SO_{2} from power-plant gases. The standard free energy of each substance are

ΔG_{f}°S(s) = 0 kJ/mol

ΔG_{f}°H_{2}O(g) = -228.57 kJ/mol

ΔG_{f}°SO_{2}(g) = -300.4 kJ/mol

ΔG_{f}° H_{2}S (g) = -33.01kJ/mol

What is the equilibrium constant for the reaction at 298K?

In principle, is this reaction a feasible method of removing SO_{2}?

If Pressure of SO_{2} = Pressure of H_{2}S and the vapor pressure of water is 26 torr, calculate the equilibrium SO _{2} pressure in the system at 298 K.

Would you expect the process to be more or less effective at higher temperatures?

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