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At 700 K the equilibrium constant for the reaction:H2(g) + I2(g) ⇌ 2HI(g)is 54.8. If 0.5 mol-1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K? |
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Answer» Let the concentration of H2(g) and I2(g) at equilibrium be x mol L-1. Then for the reaction H2(g) + I2(g) ⇌ 2HI(g) x x x Equilibrium conc. (mol/L) K = \(\frac{(0.5\, mol/L)^2}{(x\, mol/L)\times (x\, mol/L)}\) = \(\frac{0.25}{x^2}\) But K = 54.8 Therefore, \(\frac{0.25}{x^2}\) = 54.8 or, x = \(\sqrt{(0.25/54.8)}\) = 0.068 thus, at equilibrium [H2(g)] = 0.068 mol/L and [I2(g)] = 0.068 mol/L |
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