° CO 2 ( g ) + H 2 ( g ) CO( g ) + H 2 O( g ) [CO 2 ] = 0.0870 M , ° Be sure to
ID: 496202 • Letter: #
Question
°
CO2(g) + H2(g) CO(g) + H2O(g)
[CO2] = 0.0870 M,
°
Be sure to answer all parts.Consider the following equilibrium process at 686
°
C:CO2(g) + H2(g) CO(g) + H2O(g)
The equilibrium concentrations of the reacting species are [CO] = 0.0580 M, [H2] = 0.0420 M,
[CO2] = 0.0870 M,
and [H2O] = 0.0390 M.(a) Calculate Kc for the reaction at 686
°
C.(b) If we add CO2 to increase its concentration to 0.520 mol / L, what will the concentrations of all the gases be when equilibrium is reestablished?
CO2:
M
H2:
M
CO:
M
H2O:
M X
Explanation / Answer
A.) Kc = [CO]{H2O] / [CO2][H2]
Kc = 0.058x0.039 / 0.087x0.042
Kc = 0.62
B.) asssume 1L vessel
no.of moles of co2 added = 0.52
no.of moles of H2 be x
assuming hydrogen as a limiting reagent then
at equilibrium ,
no.of moles of co2 = (0.52 - x )
no.of moles of H2 = 0+0.042 =0.042
no.of moles of CO = 0.058+x
no.of moles of H2O = 0.039+x
the Kc value remains constant as there is no change in temperature.
0.62 = (0.058+x)*(0.039+x)/(0.52 -x)*(0.042)
x=0.18
concentration of CO2=0.34
conc of H2 =0.042
conc of CO = 0.238
conc of H2O = 0.219
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