H₂(g) + CO₂(g) ⇔H₂O(g) + CO(g)→ esta balanceada
Keq = {H₂O}× {CO} {H₂} × {CO₂}
Calcular las concentraciones molares (Molaridad) para cada sustancia
M = gramos / PM×litros
PM H₂ : (1 g / mol×2) = 2 g / mol
PM CO₂ : (12 g / mol×1) + (16 g / mol×2) = 12 + 32 = 44 g / mol
PMH₂O : (1 g / mol×2) + (16 g / mol×1) = 2 + 16 = 18 g / mol
Pmco
(12 g / mol×1) + 16 g / mol×1) = 12 + 16 = 28 g / mol
M H₂ = 2, 90 g ⇒ 2, 90 g = 0, 725 mol / L 2 g / mol× 2 L 4 g / mol×L
M CO₂ = 5, 038 g ⇒ 5, 038 g = 0, 05725 mol / L 44 g / mol × 2 L 88 g / mol×L
M H₂O = 1, 539g ⇒ 1, 539g = 0, 04275 mol / L 18 g / mol × 2 L 36 g / mol×L
M CO = 239, 4 g ⇒ 239, 4 g = 4, 275 mol / L 28 g / mol × 2 L 56g / mol×L
Keq = {H₂O}× {CO} ⇔ {0, 04275 mol / L}× {4, 275 mol / L} {H₂}×{CO₂} {0, 725 mol / L}× {0, 05725 mol / L}
Keq = {0, 1828 mol / L} {0, 0415 mol / L}
Keq = 4, 4.