Carbon Monoxide(0.40mol) and water(0.4 mol) are placed in a 2.0 vessel and are a

Question

Carbon Monoxide(0.40mol) and water(0.4 mol) are placed in a 2.0 vessel and are allowed to reach equilibirum at 100 degrees celsius.where the equilibrium constant k= 0.58. Calculate the equilibrium concentrations of the 4 gases: Equation: CO(g)+H2O(g)=> CO2(g)+H2(g) Kc=0.58@100degrees Celsius Using the ICE table might help but I have trouble with that alot. Thanks!

Explanation / Answer

CO + H2O ---> CO2 + H2 initial: 0.4/2 0.4/2 0 0 equilibrium: (0.4-x)/2 (0.4-x)/2 x/2 x/2 since Kc= ([CO2]*[H2]) / ([CO]*[H2O]) => 0.58 = (x^2/4) / ((0.4-x)^2/4) solving for x we get x=0.17293 mol [since other value of quadratic is negative] =>concentration at equillibrium: [CO] = (0.4-0.17293)/2 =0.1135 M [H2O] = (0.4-0.17293)/2 =0.1135 M [H2] =0.17293/2 = 0.086465 M [CO2] =0.17293/2 = 0.086465 M

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