Question

0.5 mole of hydrogen and 0.5 mole of iodine react in a 10 litre evacuated vessel at $448^0C$; hydrogen iodide is formed. The equilibrium constant Kc for the reaction is 50.
i) Calculate the number of moles of iodine which remain unreacted at equilibrium
ii) What is the value of $K_p$?

Answer 1

(a) Write the expression for the equilibrium constant:
${K_c} = \dfrac{{{{\left[ {{\text{HI}}} \right]}^2}}}{{\left[ {{{\text{H}}_2}} \right] \times \left[ {{{\text{I}}_2}} \right]}}$
(b) Use the relationship between the equilibrium constants in terms of concentration and in terms of partial pressures
${K_p} = {K_c} \times {\left( {R \times T} \right)^{\Delta n}}$

Complete step by step answer:
(a) Write the balanced chemical reaction between hydrogen and iodine to form hydrogen iodide.
${{\text{H}}_2}{\text{ + }}{{\text{I}}_2}{\text{ }} \to {\text{ 2 HI}}$
Total volume of the reaction mixture is 10 L.
Let x moles of hydrogen react to reach equilibrium

| Hydrogen| Iodine| Hydrogen iodide
---|---|---|---
Initial number of moles| 0.5| 0.5| 0.0
Change in number of moles| -x| -x| +2x
Equilibrium number of moles| 0.5-x| 0.5-x| 2x
Equilibrium concentration (M)| $\dfrac{{0.5 - x}}{{10}} = 0.05 - 0.1x$| $\dfrac{{0.5 - x}}{{10}} = 0.05 - 0.1x$| $\dfrac{{2x}}{{10}} = 0.2x$

Write the expression for the equilibrium constant and substitute values:

50 = \dfrac{{{{\left( {0.2x} \right)}^2}}}{{\left( {0.05 - 0.1x} \right) \times \left( {0.05 - 0.1x} \right)}} \\\ 50 = \dfrac{{(0.2x)^2}}{{\left( {0.05 - 0.1x} \right)^2}} \\\\$$ Take square root on both sides of equation $$\sqrt {50} = \sqrt {\dfrac{{(0.2x)^2}}{{\left( {0.05 - 0.1x} \right)^2}}} \\\ \dfrac{{0.2x}}{{\left( {0.05 - 0.1x} \right)}} = 7.07 \\\\$$ $$\left( {0.05 - 0.1x} \right) = \dfrac{{0.2x}}{{7.07}} \\\ \left( {0.05 - 0.1x} \right) = 0.02828x \\\ 0.12828x = 0.05 \\\ x = 0.39{\text{ moles}} \\\ $$ Calculate the number of moles of iodine present at equilibrium $$\left( {0.5 - x} \right) = \left( {0.5 - 0.39} \right) = 0.11{\text{ moles}}$$ (b) Calculate the value of change in the number of moles of gaseous species during the reaction $$\Delta n = 2 - \left( {1 + 1} \right) = 0$$ Use the relationship between the equilibrium constants in terms of concentration and in terms of partial pressures $${K_p} = {K_c} \times {\left( {R \times T} \right)^{\Delta n}} \\\ {K_p} = {K_c} \times {\left( {R \times T} \right)^0} \\\ {K_p} = {K_c} = 50 \\\\$$ **Note:** (a)To avoid calculation error, do not forget to raise the concentration of hydrogen iodide to the second power in the equilibrium constant expression. (b)For the reactions involving equal number of moles of gaseous reactants and products, the value of the equilibrium constant in terms of partial pressures is equal to the value of the equilibrium constant in terms of concentration.