26. Sulfur dioxide and nitrogen dioxide react at 298 K via the reaction SO2(g)+NO2(g)⇌SO3(g)+NO(g)

Question

Sulfur dioxide and nitrogen dioxide react at 298 K via the reaction

SO2(g)+NO2(g)⇌SO3(g)+NO(g)

A system at equilibrium contains 0.376 mol SO2 , 0.376 mol NO2 , 0.703 mol SO3 , and 0.695 mol NO. O2 is added to the container and NO reacts completely with the O2. Calculate the amount of SO3 in the container after the system returns to equilibrium

Answer 1

1. Calculate the K_c value. This equals the product of the moles in the products over the product of the moles in the reactants. So (0.703*0.69)/(0.376/0.376). This K_c equals 3.43.
2. NO in the products is reacted out completely. Therefore, we have to calculate the changes in moles in relation to that loss of NO. To do this, we designate NO as X, and adjust the moles by adding or subtracting Z.
3. How do we determine if we need to add or subtract? Well, since we are removing something from the products, the reaction with shift RIGHT. Meaning you are going to lose some of your reactant to gain more product and restore equilibrium.
4. The new values will be as follows: for the reactants, SO₂ and NO₂ will now be (0.376-X) in the K_c equation. SO₃ will be (0.703+X) and NO will be X.
5. The new equation to solve for X will be: 3.43=x(0.703+X)/(0.376-X)². After multiplying and reducing, you will determine that NO, or X, equals .1477 moles.
6. We want to find the change in SO_3. So, using the equation 0.703+X, 0.703+.1477=0.851 moles of SO_3.