Homework Help: College Chemistry

Posted by Morgan on Tuesday, June 12, 2012 at 1:03am.

The initial concentration for the compounds involved in the reaction shown were determined to be [CH4(g)] = 0.6616 mol/L, [H2O(g)] = 0.8127 mol/L, [CO(g)] = 0.9566 mol/L, [H2(g)] = 0.7688 mol/L. Calculate the value of the equilibrium constant (Kc) at 1200 K if the equilibrium concentration of CO(g) was 0.8900 mol/L.

CH4(g)+H2O(g) = CO(g)+3H2(g)

College Chemistry - DrBob222, Tuesday, June 12, 2012 at 2:24pm
......CH4 + H2O ==> CO + 3H2
I..0.6616.0.8127..0.9566.0.7688
C.....+x...+x......-x.....-3x
E.................0.8900.....
If initial CO is 9566 and equil CO is 0.8900, then x must be 0.9566-0.8900 = 0.0666.
Calculate CH4 = 0.6616+x for CH4, 0.8127+x for H2O, and 0.7688-3x for H2, substitute into the Kc expression and solve for Kc.

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