If 0.01 mol of #H_2# and 0.01 mol of #CO_2# are mixed in a 1 litre container at 750°C, what are the concentrations of all substances (in mol/L) present at equilibrium?

given the reaction is:
#CO_2 + H_2 harr CO + H_2O#
and the equilibrium constant, K = 0.771

Also, What is the partial pressure of #CO# in the container at equilibrium?

given the reaction is:
#CO_2 + H_2 harr CO + H_2O#
and the equilibrium constant, K = 0.771

Also, What is the partial pressure of #CO# in the container at equilibrium?

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Oct 20, 2017

Answer:

concentration of CO and #H_2O#= # 4,67 xx 10^-3 (mol)/L#
concentration of #H_2# and #CO_2#=#5,33 xx 10^-3 (mol)/L#

Explanation:

let's call x the mol of #H_2# which react (and remain at the equilibrium 0,01-x), it will be x also X the mol of #CO_2# that react (and remain at the equilibrium 0,01-x). they will form x mol of CO and x mol of water
we can write: #K_c=(X xx X)/((0,01-X)(0,01-X)) =0,771#
#X^2 = 0,771 (0,01-X)^2#
by solving the second grade equation you obtain #X= 4,67 xx 10^-3# that are the concentration in mol/L of both CO and #H_2O# while the concentration of #H_2# and #CO_2# wil be #(1-0,467) 10^2= 0,533 xx 10^(-2) = 5,33 xx 10^-3 (mol)/L#

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