# chem

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For the equilibrium reaction:
CO(g)+H2O(g)<=>CO2(g)+H2(g)
the Keq value at 690°C is 10.0. A mixture of 0.300 mol of CO, 0.300 mol of H2O, 0.500 mol of CO2 and 0.500 mol of H2 is placed in a 1.0 L flask.

a)Write the Keq expression for this reaction and use the concentrations above to calculate a trial value of the Keq. Does this value match the given Keq ? Is the reaction at equilibrium ? Explain.

b)Calculate the equilibrium concentrations of all four species.

• chem -

Keq = (CO2)(H2)/(CO)(H2O)
I assume by trial value of Keq you mean the reaction quotient. That sill be
Kquo = (0.500)(0.500)/(0.300)(0.300)
Kquo = 2.78. It doesn't match Keq of 10 and it is too low which means that the products are too small and the reactants are too large so the reaction will go to the right.

..........CO + H2O ==> CO2 + H2
initial.0.300.0.300...0.500.0.500
change...-x...-x........+x.....+x
equil..0.3-x..0.3-x..0.5+x..0.5+x

Substitute the equilibrium values of the ICE chart above and solve for x, then the individual concns of each molecule.
Post your work if you get stick. Here is a hint: When you set up the numbers in for Keq to solve the equation, note that you can take the square root of both sides and you need not solve a quadratic equation. I get x to be approximately 0.11 but that's approximate.