In the formation of acid rain, a significant reaction occurs where 2 moles of sulfur dioxide (SO₂) react with 1 mole of oxygen gas (O₂) to produce 2 moles of sulfur trioxide (SO₃). The reaction can be represented as:

2 SO₂ + O₂ ⇌ 2 SO₃

To calculate the equilibrium constant (K_c) for this reaction at 340 degrees Celsius, we start with the initial concentrations: 0.30 M of SO₂ and 0.20 M of O₂. At equilibrium, the concentration of SO₃ is found to be 0.0150 M.

We can use an ICE (Initial, Change, Equilibrium) table to track the changes in concentration. Initially, the concentrations are:

• SO₂: 0.30 M
• O₂: 0.20 M
• SO₃: 0 M

As the reaction proceeds, we denote the change in concentration of SO₂ as -2x, for O₂ as -x, and for SO₃ as +2x. Thus, the equilibrium concentrations can be expressed as:

• SO₂: 0.30 - 2x
• O₂: 0.20 - x
• SO₃: 2x

Given that at equilibrium, SO₃ = 0.0150 M, we can set up the equation:

2x = 0.0150 M

From this, we find:

x = 0.0075 M

Now, substituting x back into the expressions for the equilibrium concentrations gives us:

• SO₂: 0.30 - 2(0.0075) = 0.285 M
• O₂: 0.20 - 0.0075 = 0.1925 M
• SO₃: 0.0150 M (as given)

Next, we can set up the equilibrium constant expression:

K_c = \(\frac{[SO_3]^2}{[SO_2]^2 \cdot [O_2]}\)

Substituting the equilibrium concentrations into the expression yields:

K_c = \(\frac{(0.0150)^2}{(0.285)^2 \cdot (0.1925)}\)

Calculating this gives:

K_c = 0.01439

Considering significant figures, the final value for the equilibrium constant is reported as:

K_c = 0.014

It is important to note that equilibrium constants are dimensionless, meaning they have no units.