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
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