Question

# When N2O5(g) is heated it dissociates into N2O3(g) and O2(g) according to the reaction: N2O5(g)⇌N2O3(g)+O2(g) Kc=7.75...

When N2O5(g) is heated it dissociates into N2O3(g) and O2(g) according to the reaction:

N2O5(g)⇌N2O3(g)+O2(g)

Kc=7.75 at a given temperature

The N2O3(g) dissociates to give N2O(g) and O2(g) according the reaction:
N2O3(g)⇌N2O(g)+O2(g)

Kc=4.00 at the same temperature

When 4.00 mol of N2O5(g) is heated in a 1.00-Lreaction vessel to this temperature, the concentration of O2(g) at equilibrium is 4.50 mol/L.

Part A: Find the concentration of N2O5 in the equilibrium system.

Part B: Find the concentration of N2O in the equilibrium system.

Part C: Find the concentration of N2O3 in the equilibrium system.

irst make an ICE table (Initial concentrations, change in concentrations, equilibrium concentrations)
N2O5<--->N2O3+O2
[ ]initial: 4mol/l 0 0
[ ] change: -x +x +x
[ ] equil. 4-x x x
whats happening here is that you are showing what should happen to get the equil. since you're supposed to have 7.75 as your k at equil, and so far you don't have that (you know you dont b/c k= con. of products/ con. of reactants), your equation will favor the forward rxn. so now u have this as your equilibrium equation:
[x]*[x]/[4-x]=7.75