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Question
The value of Kc for the reaction 3O2(g)⟶2O3(g) at 25 degree C.If equilibrium concentration of O2 in air at 25 degree C is 1.6×10−2, what is the concentration of O3?
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Solution
3O2(g)=2O3(g)
equilibrium constant (Kc)=[O3]2/[O2]3
2×10−50=[O3]2/(1.6)3×10−6
[O3]2=2×10−50×(1.6)3×10−6=8.192×10−56[O3]=√8.192×10−28
[O3]=2.86×10−28M
3O2(g)=2O3(g)
equilibrium constant (Kc)=[O3]2/[O2]3
2×10−50=[O3]2/(1.6)3×10−6
[O3]2=2×10−50×(1.6)3×10−6=8.192×10−56
[O3]=√8.192×10−28
[O3]=2.86×10−28M
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2O3 (g)