Calculate the number of moles of HI that are at equilibrium with 1.39 mol of H2 and 1.39 mol of I2 in a 5.00−L flask at 448 °C. H2+I2⇌2HI Kc=50.2 at 448°C

The equilibrium moles of HI is 2.168 moles

Explanation:

We are given:

Initial moles of hydrogen gas = 1.39 moles

Initial moles of iodine = 1.39 moles

Molarity is calculated by using the formula:

Volume of the container = 5.00 L

The given chemical equation follows:

Initial:              1.39   1.39

At eqllm:      1.39-x   1.39-x    2x

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

Neglecting the value of x = 1.94 because equilibrium concentration cannot be greater than initial concentration

So, equilibrium moles of

Hence, the equilibrium moles of HI is 2.168 moles