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College Chemistry : Chemical Equilibrium, Equilibrium Constant, and Reaction Quotient

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Example Question #1 : Equilibrium

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.

Possible Answers:

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

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

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

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

Correct answer:

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

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:

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

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Example Question #1 : Equilibrium

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?

Possible Answers:

0.113

0.0121

0.250

0.158

Correct answer:

0.113

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 0.100M , 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.

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

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Example Question #2 : Equilibrium

Calculate the equilibrium constant at 298K for the reaction by using free energies of formation.

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

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

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

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

Possible Answers:

$2.39\times 10^5$

$8.55\times 10^4$

$7.44\times 10^5$

$1.95\times 10^5$

Correct answer:

$1.95\times 10^5$

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 .

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

$\ln K=12.18$

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

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Example Question #1 : Chemical Equilibrium, Equilibrium Constant, And Reaction Quotient

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

Consider a reaction mixture using the equation shown. At equilibrium the partial pressure of is 113torr and the partial pressure of $Br_{2}$ is 132torr . What is the partial pressure of NOBr in this mixture if $k_{p}=30.6$ at $298^{\circ}K$ ?

Possible Answers:

.343atm or 260.5torr

.362atm or 275torr

.308atm or 234torr

.117atm or 89.3torr

Correct answer:

.343atm or 260.5torr

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 NOBr .

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

0.343atm=P(NOBr)

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Example Question #2 : Equilibrium

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

Considering the reaction shown, if the partial pressures of , $Br_{2}$ , and NOBr are 125torr each, is the mixture at equilibrium? If not which direction will the reaction proceed to reach equilibrium if $k_{p}=30.6$ ?

Possible Answers:

No, the reaction will proceed left towards the reactants.

Yes, the reaction will move towards the products.

No, reaction will proceed right towards the products.

Yes

Correct answer:

No, reaction will proceed right towards the products.

Explanation:

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

k_p=30.6

Since k>Q 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.

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Example Question #3 : Chemical Equilibrium, Equilibrium Constant, And Reaction Quotient

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

In the laboratory .43mol of and .48mol of $H_{2}O$ are reacted in a beaker. At equilibrium .24mol of remain. Using the equation shown calculate the equilibrium constant.

Possible Answers:

$k_{c}=.28$

$k_{c}=1.93$

$k_{c}=.63$

$k_{c}=.52$

Correct answer:

$k_{c}=.52$

Explanation:

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

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

Initial .43M .48M

Change -.19M -.19M +.19M +.19M

Equilibrium .24M .29M .19M .19M

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

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

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Example Question #1 : Equilibrium

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

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 STP .

Possible Answers:

$k_{c}=.8$

$k_{c}=104.2$

$k_{c}=.0096$

$k_{c}=1.25$

Correct answer:

$k_{c}=.0096$

Explanation:

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

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

Initial .027M

Change -.012M +.012M

Equilibrium .015M .012M

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

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Example Question #3 : Equilibrium

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?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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.

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

With regards to the temperature, we will need to convert the units given in the question stem into units of Kelvin.

$T=25+273=298\:K$

Knowing that is the ideal gas constant, we have all the information we need to solve for the value of $\Delta G^{o}$ .

$\Delta G^{o}=-(8.314\:\frac{J}{mol\cdot K})(298\:K)ln(2.3\cdot 10^{-9})$

$\Delta G^{o}=49279\:\frac{J}{mol}\approx 49\:\frac{kJ}{mol}$

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Example Question #4 : Equilibrium

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

Determine the acid dissociation constant expression for the given reaction.

Possible Answers:

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

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

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

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

Correct answer:

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

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.

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College Chemistry : Chemical Equilibrium, Equilibrium Constant, and Reaction Quotient

Study concepts, example questions & explanations for College Chemistry

All College Chemistry Resources

Example Questions

Example Question #1 : Equilibrium

Example Question #1 : Equilibrium

Example Question #2 : Equilibrium

Example Question #1 : Chemical Equilibrium, Equilibrium Constant, And Reaction Quotient

Example Question #2 : Equilibrium

Example Question #3 : Chemical Equilibrium, Equilibrium Constant, And Reaction Quotient

Example Question #1 : Equilibrium

Example Question #3 : Equilibrium

Example Question #4 : Equilibrium

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