Next gasses are in balance and sit in a bottle at 980°.
CO : 0,15 atm
CO2 : 0,2 atm
H2 : 0,09 atm
H2O : 0,2 atm
a) give the equation of equilibrium
b) there is further added H2. After a new equilibrium is the pressure of CO = 0,23 atm.
What are the other partial pressures?
CO : 0,15 atm
CO2 : 0,2 atm
H2 : 0,09 atm
H2O : 0,2 atm
a) give the equation of equilibrium
b) there is further added H2. After a new equilibrium is the pressure of CO = 0,23 atm.
What are the other partial pressures?
-
CO + H2O <----> CO2 + H2
k(980) = .2 * .09/(.15 * .2) = 0.6
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the addition CO came from a loss of CO2
thus CO2 = 0.2 - (.23-.15) = .12
water gains in a similar fashion: 0.2+.08 =0.28 = H2O
to get the amount of H2 you need to go to the equil eqn
0.6 = .12 * H2/(0.23 * 0.28) solve for H2
k(980) = .2 * .09/(.15 * .2) = 0.6
===================
the addition CO came from a loss of CO2
thus CO2 = 0.2 - (.23-.15) = .12
water gains in a similar fashion: 0.2+.08 =0.28 = H2O
to get the amount of H2 you need to go to the equil eqn
0.6 = .12 * H2/(0.23 * 0.28) solve for H2