The equilibrium constant, K c, is equal to 1.4 at 1200°
K for the reaction:
CO2(g) + H2(g) ⇌ CO(g) +
H2O(g)
If 0.65 moles of CO2 and 0.65 moles of H2 are introduced into a
1.0-L flask, what will be the concentration of CO when equilibrium
is reached?
The equilibrium constant, K c, is equal to 1.4 at 1200°
K for the reaction:
CO2(g) + H2(g) ⇌ CO(g) +
H2O(g)
If 0.65 moles of CO2 and 0.65 moles of H2 are introduced into a
1.0-L flask, what will be the concentration of CO when equilibrium
is reached?
0.30 M | |
0.38 M | |
0.42 M | |
0.35 M |
we know that
conc = moles / volume
given
volume = 1 L
so
conc = moles
so
initial conc of C02 = 0.65 M
initial conc of H2 = 0.65 M
the given reaction is
C02 + H2 ---> C0 + H20
the equilibrium constant is given by
Kc = [C0] [H20] / [C02] [H2]
now
consider the reaction
C02 + H2 ---> C0 + H20
using ICE table
initial conc of C02 , H2 , C0 , H20 are 0.65 ,0.65 , 0 , 0
change in conc of C02 , H2 , C0 , h20 are -x , -x ,+x , +x
equilibrium conc of C02 , H2 , CO , H20 are 0.65 -x , 0.65 - x , x ,x
now
Kc = [C0] [H20] / [C02] [H2]
1.4 = x2 / [0.65-x]^2
solving we get
x = 0.3523
now
[CO]eq = x
[C0]eq = 0.3523
so
the conc of CO at equilibrium is 0.35 M
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