Answer :
Final answer:
To find the number of moles of HI at equilibrium, we use the initial concentrations of H2 and I2, the equilibrium constant Kc, and solve for 'x', the change in moles. The equilibrium moles of HI are found by multiplying '2x' with the volume of the flask.
Explanation:
The equilibrium calculation involves using the initial mole amounts, changes in those amounts due to the reaction reaching equilibrium, and the equilibrium constant (Kc). Given that we start with 1.35 mol of H2 and 1.35 mol of I2 in a 4.00 L flask, we denote the change in moles of HI as '2x' (since it's produced according to the stoichiometry of the balanced equation), and the decrease in moles of H2 and I2 as 'x' at equilibrium.
We set up the equilibrium expression as: Kc = [HI]2 / ([H2] [I2]), where the equilibrium concentrations ([ ]) are in moles per liter (M). After calculating concentrations by dividing moles by volume and substituting into the equilibrium expression, we solve for 'x' using the quadratic formula or a suitable method. Then, we find the moles of HI by multiplying '2x' with the volume of the flask (4.00 L).
Kc for the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 66.9 at 442°C. Since the volume of the flask is 4.00 L, the initial concentration of both H2 and I2 is 1.35 mol / 4.00 L = 0.3375 M. If 'x' is the change in moles at equilibrium, the concentration of HI produced will be 2x since the stoichiometry of the reaction produces two moles of HI for every mole of H2 and I2 reacted. The equilibrium equation is: 66.9 = (2x)2 / ((0.3375 - x) (0.3375 - x)). Upon solving for 'x', we will calculate the moles of HI by 2x times the volume of the flask (4.00 L).
