Achieve #5.57

[305769107]

The two air pollutants SO3 and NO can react as follows: SO3(g) + NO(g) --> SO2(g) + NO2(g)

(b) Given that at a certain temperature K=6.0*10^3, calculate the amount (in moles) of NO that must be added to a 1.00-L vessel containing 0.245 mol SO3 to form 0.240 mol SO2 at equilibrium.

Can someone please explain step by step how you would solve this?

[Chem_Mod]

Hi, I can't see if this K value is for Kp or Kc, but I will solve assuming the K value is Kc; the conversion from Kc to Kp is covered in textbook chapter 5H.3

Note that because the reaction vessel is 1 L, number of moles and concentration (moles/L) are identical

This is a simple ICE table problem. Notice that we are solving for the initial number of moles of NO that will get us to our desired equilibrium values. Also notice that the stoichiometry of the reaction, coupled with the fact that we initially have no SO2 or NO2, means that the equilibrium values of SO2 and NO2 will be equal. Since we are given that at equilibrium SO2 = 0.240 moles, that must also be the equilibrium value for NO2

SO3(g) + NO(g) --> SO2(g) + NO2(g)

I 0.245 ? 0 0
C -x -x +x +x
E 0.245-x ?-x 0.240 0.240

We have now established that x is 2.40. That means at equilibrium, SO3 = 0.005 moles. We can then set up our equilibrium expression, 6.0 x 10³ = [(.240)(.240)/(0.005)(? - 0.240)]. You can work out the algebra to get to the final answer from here.