# Unlocking the Equilibrium Equation: Calculating Grams of HI

## How can we calculate the number of grams of HI at equilibrium?

Given the equation H_{2} + I_{2} ⇌ 2HI with K_{c} = 5.02x10^{-4} at 448°C, how do we determine the grams of HI in equilibrium with 1.25 mol of H_{2} and 63.5 g of iodine?

## Calculating the Equilibrium Concentration of HI

To calculate the number of grams of HI at equilibrium, we need to follow a few steps. Firstly, we convert the moles of H_{2} and iodine to concentrations using the ideal gas law. Next, utilizing stoichiometry, we determine the equilibrium concentration of HI. Finally, we convert the concentration to grams using the molar mass of HI.

When tackling equilibrium calculations like this, it's essential to approach the problem systematically. Begin by converting the given moles of H_{2} and the mass of iodine to concentrations in mol/L using the ideal gas law. This step allows us to establish the initial concentrations of the reactants.

Once we have the initial concentrations of H_{2} and iodine, we can proceed to determine the equilibrium concentration of HI. By applying stoichiometry based on the balanced equation, we can find the molar ratios between H_{2}, I_{2}, and HI at equilibrium.

After finding the equilibrium concentration of HI, the final step involves converting this concentration to grams. This conversion is achieved by utilizing the molar mass of HI, which allows us to relate the concentration in mol/L to grams.

By following these steps methodically, you can successfully calculate the grams of HI at equilibrium in a chemical reaction. It's a process that combines the principles of stoichiometry, equilibrium constants, and molar conversions to determine the final result accurately.