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Equilibrium...?
You place 3.00 mol of pure SO3 in an 8.00 L flask at 1150 K. At equilibrium, 0.58 mol of O2 has been formed. Calculate K for the reaction at 1150 K.
2SO3(g) ↔ 2SO2(g) + O2(g)
show work plz.
2 Answers
- Dr.ALv 71 decade agoFavorite Answer
Initial concentration SO3 = 3.00 mol / 8.00L = 0.375 M
Concentration O2 at equlibrium = 0.58 mol / 8.00 = 0.0725 M
2 SO3 <-----> 2SO2 + O2
initial concentration
0.375
change
-2x. . . . . . . . .. +2x. . . .+x
at equilibrium
0.375-2x . . . .. . 2x. . . . .. x
x = 0.0725
2x = 0.145
concentration SO3 = 0.375 - 0.145 = 0.230 M
K = (0.145)^2 ( 0.0725) / (0.230)^2 = 0.0288
- pisgahchemistLv 71 decade ago
The equilibrium constant is computed from the equilibrium concentrations.
Start with the balanced chemical equation
2SO3(g) <==> 2SO2(g) + O2(g)
Compute the equilibrium concentrations of the reactants and products.
[SO3] = 2.42 mol / 8 L = 0.3025 mol / L
[SO2] = 0.58 mol / 8 L = 0.0725 mol / L
[O2] = 0.29 mol / 8 L = 0.03625 mol/L
Plug the equilibrium concentrations into the equation, and don't forget to square the ones that need it.
Keq = [SO2]^2 [O2] / [SO3]^2
Keq = 0.00208
Source(s): 35 years of teaching chemistry
