Question: Calculate 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.

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.needs to be calculated

The equilibrium mixture has equal rates of the forward and backward (reverse) reactions.

For a chemical reaction as follows:

\(A \to B + C\)

The expression for equilibrium constant will be:

\({K_C} = \frac{{\left( B \right)\left( C \right)}}{{\left( A \right)}}\)

Here, [A], [B] and [C] is equilibrium concentration of A, B and C respectively.

The number of moles (n) of I₂ can be calculated from its mass(m) and molar mass (M) as follows:

\(n = \frac{m}{M}\)

Putting the values,

\(\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,n = \frac{{63.5{\rm{g}}}}{{253.809{\rm{gmo}}{{\rm{l}}^{ - 1}}}}}\\{ = 0.250{\rm{mol}}}\end{array}\)

Expression for equilibrium constant is as follows:

\({K_c} = \frac{{{{\left( {HI} \right)}^2}}}{{\left( {{H_2}} \right)\left( {{I_2}} \right)}}\)

Now,

\(\begin{array}{l}{\left( {HI} \right)^2} = {K_c} \times \left( {{H_2}} \right) \times \left( {{I_2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = 50.2 \times 1.25{\rm{mol}} \times 0.250{\rm{mol}}\\\,\,\,\,\,\,\,\,\,\,\,\, = 15.7\,mo{l^2}\end{array}\)

\(HI = \sqrt {15.7{\rm{mo}}{{\rm{l}}^2}} = 3.96{\rm{mol}}\)

\(Mass({\rm{Hl}}) = 3.96{\rm{mol}} \times 127.9124{\rm{g}}/{\rm{mol}} = 507{\rm{g}}\)