Question

The $ Kp $ value for the reaction: $ {{H}_{2}}+{{I}_{2}}2HI $
 At $ 460{}^\circ C $ is $ 49. $ If the initial pressure of $ {{H}_{2}} $ and $ {{I}_{2}} $ is $ 0.5 $ atm respectively, determine the partial pressure of each gas at equilibrium.

Answer 1

Hint :We know that to answer this question, you must recall the formula of equilibrium constant of a reaction in terms of the concentration of reactants, that is $ Kp. $ It is given by the concentration at equilibrium of products divided by the concentration at equilibrium of reactants raised to the power of their respective stoichiometric coefficients.

Complete Step By Step Answer:
While dealing with the degree of dissociation, and number of moles, you can assume the initial number of moles of reactants to be one. This will lead to an easier calculation. You should take care of the stoichiometry of the reaction while dealing with the degree of dissociation. Also the numerical value of equilibrium constant is obtained by allowing a reaction to proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. The concentrations are measured at equilibrium; the equilibrium constant remains the same for a given reaction independent of initial concentrations of the reactants and products. This knowledge allows us to derive a model expression that can serve as a standard for any reaction.

	$ {{H}_{2(g)}} $	$ {{I}_{2}} $	$ 2H{{I}_{(g)}} $
Initial (atm) Pressure	$ 0.5 $	$ 0.5 $	$ 0 $
Final (atm) Pressure	$ 0.5-x $	$ 0.5-x $	$ 2X $

Thus, $ Kp=\dfrac{{{\left( pHI \right)}^{2}}}{\left( p{{H}_{2}} \right)\left( p{{I}_{2}} \right)} $
On substituting the values we get;
 $ 49=\dfrac{{{\left( 2X \right)}^{2}}}{{{\left( 0.5-X \right)}^{2}}}\Rightarrow 7=\dfrac{\left( 2X \right)}{\left( 0.5-X \right)} $
On further solving we get:
 $ 3.5\times \left( -7X \right)=2X\Rightarrow X=\dfrac{3.5}{9}=0.389 $
Thus, $ p{{H}_{2}} $ at equilibrium $ =\left( 0.5-0.389 \right)=0.111 $
We should remember the relation between different equilibrium constants while doing these questions. These relations help us in solving these questions with less difficulty. Always remember the expressions for equilibrium constants depend on the equilibrium values rather than initial values. The expression for will depend on the equilibrium partial pressure of all the species rather than depending on the initial values.

Note :
Remember that always remember that as two or more molecules of any gas are produced in the reaction, then while calculating the decrease or increase in their partial pressure make sure that you multiply the increase or decrease by the number of molecules present. e.g. We have multiplied the decrease in partial pressure of sulfur trioxide gas here by two as its two molecules are there.