"Published in Newark, California, USA"
For the reaction
at 500°C, Keq = 62.5. If 5.0 moles of H2 and 5.0 moles of I2 are placed in a 10L container at 500°C and allowed to come to equilibrium, calculate the final concentration of H2, I2, 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 H2 that will react with I2 to form HI. Since the coefficient of H2 and I2 are the same, then the number of moles of I2 that will react with H2 to form HI is also x.
Let 2x be the number of moles of HI that is formed by the reaction of H2 and I2. Because of the stoichiometric relationship of H2, I2, and HI, then x is the number of moles of H2, x is the number of moles of I2, 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], [H2], and [I2] 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:
