Chemistry Le Chat question

Question

SO₂(g) + NO₂(g)  SO₃(g) + NO (g)

At a given temperature, analysis of an equilibrium mixture found [SO₂] = 4.00 M, [NO₂] = 0.500 M, [SO₃] = 3.00 M, and [NO] = 2.00 M.

(a) What is the new equilibrium concentration of NO when 1.50 moles of NO₂ are added to the equilibrium mixture? Assume a 1.00 L container. 

(b) How many moles/liter of NO₂ would have to be added to the original equilibrium mixture to increase the equilibrium concentration of SO₃ from 3.00 M to 4.20 M at the same temperature? 

Answer 1

First, find Keq from the equilibrium concentrations provided.

Keq = [SO₃][NO]/[SO₂][NO₂] = (3.00)(2.00)/(4.00)(0.500) = 3.00

Next, set up an ICE table reflecting conditions described:

SO2 + NO2 <===> SO3 + NO

4..........0.5................3............2.........I

-x........-x.................+x..........+x........C

4-x....0.5-x............3+x.........2+x.......E

Keq = 3.00 = (3+x)(2+x)/(4-x)(0.5-x)

Solve for x. I found it to be x = 1.7 (please do check my math. I'm better at chemistry than at math).

Assuming this is correct, then plug this in to the E in the table where NO = 2+x and that means that

new concentration of NO = 2 + 1.7 = 3.7 M

For part (b), take a similar approach but you will use the initial concentrations and the final equilibrium concentration of SO3 will be 4.20, meaning it increased by 1.20 M. Hopefully you can figure this out.