For the equilibrium reaction 2so2 (g) + o2 (g) 2so3 (g), δhºrxn = - 198 kj/mol. which one of these factors would cause the equilibrium constant to increase?

Question

Ronin Haas

Chemistry

29 April, 05:19

For the equilibrium reaction 2so2 (g) + o2 (g) 2so3 (g), δhºrxn = - 198 kj/mol. which one of these factors would cause the equilibrium constant to increase?

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Camron Garza · Answer 1

Let's rewrite the reaction for clarity:

2 SO₂ (g) + O₂ (g) ⇆ 2 SO₃ (g) δhºrxn = - 198 kj/mol

The equilibrium constant of a reaction is the ratio of the concentration its products to its reactants which are raised to their respective stoichiometric coefficients. For this reaction, the K would be

K = [SO₃]²/[SO₂]²[O₂]

To get a larger K, the products must be greater than the reactants. This means that the forward reaction must be favored to yield more of the product SO₃. There are different ways to do this: by manipulating the pressure, concentration or temperature.

For the concentration, you should add more amounts of the reactants. For the pressure, we should increase it. This is because the product side has only 2 moles of gas compared to 3 moles of gas in the reactants. So, it wall have more room for the product even at a higher pressure. Lastly, since the reaction is exothermic manifested by the negative sign of δhºrxn, the reaction would favor the forward reaction at high temperatures.