# 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.