What is the equilibrium concentration of water in the reaction #"CO" + "H"_2 ⇌ "CO"_2 + "H"_2"O"# if the initial concentrations of carbon monoxide and hydrogen are each 0.125 mol/L?

Question

939 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-28T04:59:35

without the temperature of the reaction, it is impossible solve the problem. There are necessary also the thermodynamic data for enthalpy and entropy

Answer 2 · 2015-03-31T21:21:44

At 550 K, the equilibrium concentration of H₂O is 0.015 mol/L.

You don't give a value of #K#, so I will arbitrarily pick a number.

At 550 K, #K = 58#.

First, you write the chemical equation with an ICE table.

enter image source here

The equilibrium constant expression is

#K = "[CO][H₂O]"/"[CO₂][H₂]" = (x × x)/((0.125-x)(0.125-x)) = x^2/(0.125-x)^2 = 58#

#x/(0.125 – x) = 7.62#

#x = 7.62(0.125-x) = 0.952 -7.62x#

#8.62x = 0.952#

#x = 0.110#

#"[H₂O]" = (0.125 – x)" mol/L" = "(0.125 – 0.110) mol/L" = "0.015 mol/L"#

Check: #0.110^2/0.015^2 = 54#. Close enough to 2 significant figures!