For the reaction given below at 700 degrees C , Kc = 0.534 H2(g) + CO2(g) <-> H2

Question

For the reaction given below at 700 degrees C , Kc = 0.534 H2(g) + CO2(g) <-> H2O(g) + CO(g) Calculate the number of moles of H2 that are present at equilibrium if a mixture of 0.245 miles of CO and 0.245 moles of H2O is heated to 700 degrees C in an 11.4 L container. I know to solve for the concentrations of CO and H2O by dividing # of moles by the volume to get Molarity. [CO] = 0.0215 [H2O] = 0.215 My problem is when solving for the concentration of H2 there is a CO that I don't have a concentration for. [CO][H2O] / [H2][CO] Is the CO not supposed to be there? How do I finish solving this problem?

Explanation / Answer

H2(g) + CO2(g) H2O(g) + CO(g) Kc = 0.534 = (0.245-X)^2 / X^2 => X = 0.142 moles = 0.142 / 11.4 = 0.0124 M of CO2 = For H2

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