Here is an equilibrium: H2(g) + CO2(g) ? H2O(g) + CO(g) K = 1.6 If initially the

Question

Here is an equilibrium:
H2(g) + CO2(g) ? H2O(g) + CO(g) K = 1.6
If initially the concentrations of H2 and CO2 were 0.122 mol/L and nothing else was present, what will be the concentrations of H2O and CO at equilibrium?


PS can you show all of your mathematical steps bc thats what im having trouble with?

Explanation / Answer

H2(g) + CO2(g) = H2O (g) + CO(g) k = [H2O] [CO] / [H2] [CO2] Q = [.75] [1] / [1] [2] = 0.375 ,....Q is too small compared to the K, therefor the reaction will shift to increase the numerator (toward the products) I realize that I am not dividing the four sets of moles by 5 litres to change the moles into molarities, but that effect would all cancel out & still give the same answer ============= .... .....H2(g) + CO2(g) H2O (g) + CO(g) was ,,,]1.00 .. ... 2.00.... ... 0.75 ... ...1.00 Shifts..(1.00-x) ...(2.00-x).... => (0.75+x) ... (1.00+x) K = [H2O] [CO] / [H2] [CO2] 1.6 = [.75+x] [1.00+x] / [1.00-x] [2.00-x] 1.6 = (0.75 +1.75x +x2) / (2.00 -3.00x +x2) 1.6 (2.00 -3.00x +x2) = (0.75 +1.75x +x2) 3.2 - 4.8x + 1.6x2 = (0.75 +1.75x +x2) 0.6x2 - 6.55x +2.45 = 0 using the quadratic @ http://www.math.com/students/calculators… X = 0.388 moles shift ======================= H2O =0.75 moles + x = 0.75 moles +0.388 moles = your answer: 1.14 moles of H2O a person could go back & divide all four sets of moles by 5 litres at the beginnign too get molarities, then multiply by 5 liters at the end to return to moles, but you will still get 1.14 moles H2O

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