The equilibrium constant for the gas phase reaction of CO with water to form CO2
ID: 816551 • Letter: T
Question
The equilibrium constant for the gas phase reaction of CO with water to form CO2 and molecular hydrogen is .58 at 1000 degrees. If a mixture of 0.0200 molar CO, .0100 molar H2O and .0050 molar CO2 is allowed to come to equilibrium. What will the equilibrium ce creations of all four species be? Please show all steps thanks! The equilibrium constant for the gas phase reaction of CO with water to form CO2 and molecular hydrogen is .58 at 1000 degrees. If a mixture of 0.0200 molar CO, .0100 molar H2O and .0050 molar CO2 is allowed to come to equilibrium. What will the equilibrium ce creations of all four species be? Please show all steps thanks!Explanation / Answer
CO + H2O ----> CO2 + H2
As the equilibrium constant is:
K = [Products]n / [Reagents]n where n is the stoichiometric coefficient = ( [CO] * [H2O] ) / ( [CO2] * [H2] )
Now, as [M] = moles / volume and the volume is the same for every component, you'll notice that the volume is irrelevant in the equation because it is annulled between each other.
K = moles CO * moles H2O / moles CO2 * moles H2
Ok, then let's write the reaction. Notice that the limitant reagent is water becase the rate is 1:1 and there is less water than CO.
CO + H2O ----> CO2 + H2
initial: 0.02 0.01 0.005 -
change -x -x +x +x
end 0.02-x 0.01-x 0.005+ x x
K = (x)*(0.005 + x) / (0.02-x)*(0.01 -x) solve for x
K * (0.02-x)*(0.01 -x) = 0.005 x + x2
K * (0.0002 -0.02 x - 0.01 x + x2) =0.005 x + x2
0.58 * (0.0002 -0.03 x + x2) = 0.005 x + x2
0.00012 - 0.174 x - 0.005 x + 0.58 x2 - x2 = 0
0.00012 - 0.169 x -0.42 x2 = 0
solve for "x"
x= 7.09 E-4 moles
moles CO eq = 0.02 - 7.09E-4 = 0.0193
moles H2O eq = 0.01 - 7.09E-4 = 0.0093
moles CO2 eq = 0.005 - 7.09 E-4 = 0.0043
moles H2 eq = 7.09 E-4
final moles CO =
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