Here we're told that an important reaction in the formation of acid rain is 2 moles of sulfur dioxide reacts with 1 mole of oxygen gas to produce 2 moles of sulfur trioxide gas. Initially, 0.30 Molar of sulfur dioxide and 0.20 Molar of oxygen are mixed and allowed to react in an evacuated flask at 340 degrees Celsius. When equilibrium is established, the equilibrium amount of sulfur trioxide was found to be 0.0150 Molar. We need to calculate the equilibrium constant k_c for the reaction at 340 degrees Celsius. So, we'll follow steps 1 to 4 to determine the equilibrium constant expression for the reaction. Remember, ICE stands for Initial, Change, and Equilibrium.

Initially, we have 0.30 Molar sulfur dioxide, 0.20 Molar oxygen, and no products, so initially, it's 0. During the reaction, the changes for the reactants are -2x for sulfur dioxide and -x for oxygen, and for the product sulfur trioxide, it's +2x, due to their stoichiometric coefficients. Therefore, at equilibrium, the concentrations are:

• Sulfur dioxide: 0.30 - 2x
• Oxygen: 0.20 - x
• Sulfur trioxide: 0 + 2x = 0.0150

We set up our equilibrium constant expression k_c, which is based on the products over reactants:

k_c = [SO 3 2 [SO 2 2 ⁢ [O 2 ]

Given the equilibrium amount of sulfur trioxide, we set the calculated 2x equal to the known equilibrium concentration which is 0.0150 Molar. Solving for x:

x = 0.0075

Using this value, we can calculate the equilibrium concentrations:

• Sulfur dioxide: 0.30 - 2(0.0075) = 0.285 M
• Oxygen: 0.20 - 0.0075 = 0.1925 M
• Sulfur trioxide: 2x = 0.0150 M (already given)

Substituting these values into the equilibrium expression:

0.0150 2 ( 0.285 2 )( 0.1925 ) = 0.01439

Rounding to match the least number of significant figures given in the reactants (2 significant figures), k_c = 0.014 (dimensionless).

This will be our final answer, indicating the equilibrium constant at 340 degrees Celsius for this reaction in the production of sulfur trioxide.