AP Chemistry - Chemical Equilibrium, Equilibrium Constant, and Reaction Quotient

Consider the following reaction:

$4\text{HCl}(g)+\text{O}_2(g)\rightleftharpoons 2\text{H}_2\text{O}(l)+2\text{Cl}_2(g)$

The reaction mixture at $850 \degree C$ initially contains $[\text{HCl}]=0.750M$ and $[\text{O}_2]=2.00M$ . At equilibrium, $[\text{HCl}]=0.650M$ . What is the equilibrium constant for the reaction?

Explanation

Start by writing the equilibrium expression:

$K=\frac{[Cl_2]^2}{[HCl]^4[O_2]}$

Now, create a chart like the following to keep track of the changes in concentration.

| | | | | | | ------------------------------------------------------------------------------------------------------------- | ---------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------- | ----- | | Initial | 0.750 | 2.00 | 0.00 | | Change | -0.100 | -0.025 | 0.200 | | Equilibrium | 0.650 | 19.75 | 0.200 |

Since we know that the concentration of HCl decreased by , we can use the stoichiometric ratios to deduce the amount of change for the oxygen gas and the chlorine gas.

Plug in the equilibrium concentrations into the expression for the equilibrium constant.

$K=\frac{(0.200)^2}{(0.650)^4(1.975)}=0.113$

$2NO(g)+Br_{2}(g)\rightleftharpoons 2NOBr(g)$

Consider a reaction mixture using the equation shown. At equilibrium the partial pressure of is and the partial pressure of $Br_{2}$ is . What is the partial pressure of in this mixture if $k_{p}=30.6$ at $298^{\circ}K$ ?

Explanation

$k_{p}= \frac{P(product)^{a}}{P(reactant)^{b}*P(reactant)^{c}}$

$k_{p}=30.6= \frac{P(NOBr)^{2}}{(\frac{132}{760})*(\frac{113}{760})^{2}}$

Use algebra to solve for the partial pressure of .

$\sqrt{30.6*(\frac{132}{760})*(\frac{113}{760})^{2}}=P(NOBr)$

$2NO(g)+Br_{2}(g)\rightleftharpoons 2NOBr(g)$

Considering the reaction shown, if the partial pressures of , $Br_{2}$ , and are each, is the mixture at equilibrium? If not which direction will the reaction proceed to reach equilibrium if $k_{p}=30.6$ ?

No, reaction will proceed right towards the products.

Yes

No, the reaction will proceed left towards the reactants.

Yes, the reaction will move towards the products.

Explanation

$Q_{p}=\frac{(\frac{125}{760})^{2}}{(\frac{125}{760})*(\frac{125}{760})^{2}}=6.08$

Since the reaction is not at equilibrium. This means that at equilibrium, the ratio of products to reactants is greater than at the given conditions. Thus, the reaction will move right towards the products to reach equilibrium.

$CO(g)+H_{2}O(g)\rightleftharpoons CO_{2}(g)+H_{2}(g)$

In the laboratory of and of $H_{2}O$ are reacted in a beaker. At equilibrium of remain. Using the equation shown calculate the equilibrium constant.

$k_{c}=.52$

$k_{c}=.63$

$k_{c}=1.93$

$k_{c}=.28$

Explanation

Use an ice table and the $k_{c}$ equation to solve.

$H_{2}O$ $CO_{2}$ $H_{2}$

Initial

Change

Equilibrium

$k_{c}= \frac{[H_{2}][CO_{2}]}{[CO][H_{2}O]}$

$k_{c}=\frac{.19M*.19M}{.24M*.29M}=.52$

$N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)$

Find the $k_{c}$ of the reaction if you start with $.027M N_{2}O_{4}(g)$ and end with $.015M N_{2}O_{4}(g)$ at .

$k_{c}=.0096$

$k_{c}=104.2$

$k_{c}=.8$

$k_{c}=1.25$

Explanation

Use an ICE table and the $k_{c}$ equation to solve.

$[N_{2}O_{4}]$ $[NO_{2}]$

Initial

Change

Equilibrium

$k_{c}= \frac{[NO_{2}]^2}{[N_{2}O_{4}]}=\frac{.012M^2}{.015M}=.0096$

$HCO_{3}^{-}(aq)\ +\ H_{2}O(l)\rightleftharpoons\ H_{3}O^{+}(aq)\ +\ CO_{3}^{2-}(aq)$

Determine the acid dissociation constant expression for the given reaction.

$K_{a}=\frac{[H_{3}O^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}]}$

$K_{a}=\frac{[HCO_{3}^{-}]}{[H_{3}O^{+}][CO_{3}^{2-}]}$

$K_{a}=\frac{[HCO_{3}^{-}][H_{2}O]}{[H_{3}O^{+}][CO_{3}^{2-}]}$

$K_{a}=\frac{[H_{3}O^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}][H_{2}O]}$

Explanation

Acid dissociation constant which is denoted as $K_{a}$ is the equilibrium constant for the ionization of an acid. Therefore, the numerator contains the product of the concentrations of the substances on the product side of the chemical equation. The denominator contains the product of the concentrations of the substances on the reactant side of the chemical equation. $H_{2}O$ is omitted in the acid dissociation constant expression because as the solvent it is in excess and therefore the change in its concentration is negligible in comparison to the other substances in solution.

Consider the following reaction:

$2\text{Si}_2\text{H}_6(g)+7\text{O}_2(g)\rightleftharpoons4\text{SiO}_2(g)+6\text{H}_2\text{O}(l)$

Give the expression for the equilibrium constant for this reaction.

$\frac{(\text{P}_{SiO_2})^4}{(\text{P}_{Si_2H_6})^2(\text{P}_{O_2})^7}$

$\frac{(\text{P}_{SiO_2})^4[(\text{H}_2\text{O}])^6}{(\text{P}_{Si_2H_6})^2(\text{P}_{O_2})^7}$

$\frac{(\text{P}_{Si_2H_6})^2(\text{P}_{O_2})^7}{(\text{P}_{SiO_2})^4}$

$(\text{P}_{Si_2H_6})^2(\text{P}_{O_2})^7(\text{P}_{SiO_2})^4$

Explanation

Recall how to find the expression of the equilibrium constant for the simplified equation:

$aA+bB\rightleftharpoons cC+dD$

$\text{K}_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Since the given equation has gases, we will only consider the partial pressures of each gas in the expression for the equilibrium constant. Remember that only molecules in aqueous and gas forms are included in this expression. Pure solids and pure liquids are excluded.

Thus, we can then write the following equilibrium constant for the given equation:

$\text{K}_{eq}=\frac{(\text{P}_{SiO_2})^4}{(\text{P}_{Si_2H_6})^2(\text{P}_{O_2})^7}$

Calculate the equilibrium constant at for the reaction by using free energies of formation.

$I_2(g)+Cl_2(g)\rightleftharpoons 2ICl(g)$

$\Delta G^{\degree}_{f} \text{ of I}_2(g)=19.3\frac{kJ}{mol}$

$\Delta G^{\degree}_{f} \text{ of Cl}_2(g)=0\frac{kJ}{mol}$

$\Delta G^{\degree}_{f} \text{ of ICl}(g)=-5.44 \frac{kJ}{mol}$

$1.95\times 10^5$

$2.39\times 10^5$

$8.55\times 10^4$

$7.44\times 10^5$

Explanation

Start by using the free energies of formation to find $\Delta G^{\degree}_{rxn}$ .

$\Delta G^{\degree}_{rxn}=\Sigma n\Delta G^{\degree}_f \text{ of products}-\Sigma n\Delta G^{\degree}_f \text{ of reactants}$ ,

$\Delta G^{\degree}_{rxn}=2(-5.44)-19.3=-30.18\text{ kJ}$

Recall the equation that links together $\Delta G^{\degree}_{rxn}$ with the equilibrium constant, .

$\Delta G^{\degree}_{rxn} =-RT \ln K$

Plug in the given information and solve for .

$-30.18(1000)=-(8.314)(298)\ln K$

$\ln K=12.18$

$K=e^{12.18}=1.95\times 10^5$

Hypobromous acid will dissociate in water at $25^{o}C$ with a $K_{a}=2.3\cdot10^{-9}$ . What is the $\Delta G^{o}$ for this dissociation process?

$+49:\frac{kJ}{mol}$

$+55:\frac{kJ}{mol}$

$+33:\frac{kJ}{mol}$

$-49:\frac{kJ}{mol}$

$-33:\frac{kJ}{mol}$

Explanation

For this question, we're given the acid dissociation constant for a reaction that occurs at a given temperature. We're asked to find the standard free energy change for the reaction.

First, we're going to need to use an equation that relates standard free energy changes with an equilibrium constant, which is shown as follows.

$\Delta G^{o}=-RTln(K)$