Calculate the Number of Grams of HI at Equilibrium

How can we determine the number of grams of HI that are at equilibrium?

Given the equilibrium constant (Kc) of 50.2 at 448°C, along with 1.25 mol of H₂ and 63.5 g of iodine, how do we calculate the grams of HI at equilibrium?

Calculating the Number of Grams of HI at Equilibrium

To determine the number of grams of HI that are at equilibrium with 1.25 mol of H₂ and 63.5 g of iodine at 448°C, we can use the equilibrium constant (Kc) and stoichiometry of the reaction.

Equilibrium calculations involve understanding the balance between reactants and products in a chemical reaction. In this case, we are given the initial amounts of H₂ and iodine (I₂) and asked to find the grams of hydrogen iodide (HI) at equilibrium.

First, we need to determine the initial moles of H₂ and I₂. With 1.25 mol of H₂ and 63.5 g of iodine, we can convert the grams of iodine to moles using its molar mass of 253.8 g/mol. This gives us approximately 0.25 mol of iodine.

Next, we utilize the stoichiometry of the reaction to analyze the change in moles for each substance. From the balanced equation H₂ + I₂ → 2HI, we determine that 1 mol of H₂ reacts with 1 mol of I₂ to produce 2 mol of HI.

Using the equilibrium constant (Kc) expression Kc = [HI]² / ([H₂][I₂]), we substitute the known values to find the concentration of HI at equilibrium. By solving for [HI], we calculate that approximately 2.52 mol/L of HI is present.

Finally, we convert the moles of HI to grams using its molar mass of 33.98 g/mol. Multiplying 2.52 mol/L by 33.98 g/mol, we determine that around 94.38 grams of HI are at equilibrium under the specified conditions at 448°C.

Therefore, the number of grams of HI at equilibrium with 1.25 mol of H₂ and 63.5 g of iodine at 448°C is approximately 94.38 grams.