Consider the following equilibrium process at 686C: CO 2(g) + H 2(g) CO (g) + H

Question

Consider the following equilibrium process at 686C: CO2(g) + H2(g) CO(g) + H2O(g) The equilibrium concentrations of the reactingspecies are [CO] = 0.050M, [H2] = 0.045M,[CO2] = 0.086M, and [H2O] = 0.040M. (a)Calculate Kc for the reaction at 686C. (b) If weadd CO2 to increase its concentration to 0.60 mol/L,what will the concentrations of all the gases be whenequilibrium is reestablished?


I Have gotten about half way through this problem, but I keepgetting stuck! PLEASE HELP. I'm pulling my hair outover here!
Consider the following equilibrium process at 686C: CO2(g) + H2(g) CO(g) + H2O(g) The equilibrium concentrations of the reactingspecies are [CO] = 0.050M, [H2] = 0.045M,[CO2] = 0.086M, and [H2O] = 0.040M. (a)Calculate Kc for the reaction at 686C. (b) If weadd CO2 to increase its concentration to 0.60 mol/L,what will the concentrations of all the gases be whenequilibrium is reestablished?


I Have gotten about half way through this problem, but I keepgetting stuck! PLEASE HELP. I'm pulling my hair outover here!

Explanation / Answer

                  CO2(g)+ H2(g) CO(g) +H2O(g) Kc = [CO] [H2O] / { [ CO2] [H2] }       = 0.05 * 0.04 / { 0.086 * 0.045}       = 0.517 CO2 concentration increased to 0.60 mol/L. So the reactionwill shift towards the products side.                      CO2(g)+ H2(g)       CO(g) +H2O(g) Initial            0.60      0.045            0.05         0.04 Change            -x         -x                  +x            +x equilibrium    (0.6 -x) (0.045-x)   (0.05 +x)   (0.04 +x) Kc = 0.517 = (0.05+x)(0.04+x) / { (0.6 -x) (0.045-x) } Solving ,we get x = 0.0273 So at the new equilibrium, [CO2] = 0.6 - 0.0273 = 0.5727M                                        [H2]= 0.045 - 0.0273 = 0.0177 M                                        [CO]= 0.05 + 0.0273 = 0.0773 M                                        [H2O]= 0.04 + 0.0273 = 0.0673 M              Initial            0.60      0.045            0.05         0.04 Change            -x         -x                  +x            +x equilibrium    (0.6 -x) (0.045-x)   (0.05 +x)   (0.04 +x) Kc = 0.517 = (0.05+x)(0.04+x) / { (0.6 -x) (0.045-x) } Solving ,we get x = 0.0273 So at the new equilibrium, [CO2] = 0.6 - 0.0273 = 0.5727M                                        [H2]= 0.045 - 0.0273 = 0.0177 M                                        [CO]= 0.05 + 0.0273 = 0.0773 M                                        [H2O]= 0.04 + 0.0273 = 0.0673 M             

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