__Category__: Chemical Engineering Math, Algebra"Published in Newark, California, USA"

For the reaction

at 500°C, K

_{eq}= 62.5. If 5.0 moles of H

_{2}and 5.0 moles of I

_{2}are placed in a 10L container at 500°C and allowed to come to equilibrium, calculate the final concentration of H

_{2}, I

_{2}, and HI.

__Solution__:

Consider the given chemical reaction above

Start 5.0 5.0 0

End 5.0 - x 5.0 - x 2x

Let x be the number of moles of H

_{2}that will react with I

_{2}to form HI. Since the coefficient of H

_{2}and I

_{2}are the same, then the number of moles of I

_{2}that will react with H

_{2}to form HI is also x.

Let 2x be the number of moles of HI that is formed by the reaction of H

_{2}and I

_{2}. Because of the stoichiometric relationship of H

_{2}, I

_{2}, and HI, then x is the number of moles of H

_{2}, x is the number of moles of I

_{2}, and 2x is the number of moles of HI.

__Note__: It is important to balance the chemical reactions first all the time in order to assign the number of moles of the reactants and products correctly either at the start or at the end of the reaction.

Since the given chemical reaction is equilibrium or reversible reaction, then the equilibrium constant is calculated as follows

where [HI], [H

_{2}], and [I

_{2}] are the concentration of reactants and products in molarity or moles of solute per liter of solution.

Calculate the concentration of [HI], [H2], and [I2] as follows:

Substitute the above values to the above equation, we have

Take the square root on both sides of the equation, we have

Therefore, the final concentration of the product:

Therefore, the final concentration of the reactants: