## 11.117

Michelle_Nguyen_3F
Posts: 40
Joined: Wed Sep 21, 2016 2:59 pm

### 11.117

The question reads: "The two air pollutants SO3 and NO can react in the atmosphere as follows: $SO_{3} (g) + NO(g)\rightarrow SO_{2}(g) + NO_{2}(g)$. b) Give that a t a certain temperature K = 6.0 x 10^3, calculate the amount (in moles) of NO that must be added to 1.00 L vessel containing 0.245 mol SO3(g) to form 0.240 mol SO2 (g). Why is it that the the solution manual says that at equilibrium, SO3=0.245 moles - 0.240 moles = 0.005 moles? Thank you!

Allison Suzuki 2B
Posts: 26
Joined: Wed Sep 21, 2016 2:57 pm

### Re: 11.117

In the problem, you're given two values: 0.245 mol SO3 and 0.240 mol SO2. The volume is 1.00 L, so your concentration will be of the same values. When you create your ICE table, the reaction initially starts off with none of the product (SO2), and ends up with 0.240, you can conclude that the change in concentration, +x, equals 0.24.
Going back to SO3, the change will be -x, and since x=0.24, you can conclude that the concentration will be 0.245-0.240 = 0.005

Hope this helps!

Paula Dowdell 1F
Posts: 72
Joined: Tue Nov 15, 2016 3:00 am

### Re: 11.117

For the same reasons NO is x-0.240