Equilibrium - College Chemistry

Card 0 of 17

Question

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.

Answer

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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Question

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

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 $298^{\circ}K$ ?

Answer

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

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

Use algebra to solve for the partial pressure of .

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

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Question

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

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 $k_{p}=30.6$ ?

Answer

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

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.

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Question

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

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.

Answer

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

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

Initial

Change

Equilibrium

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

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

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Question

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

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 .

Answer

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

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

Initial

Change

Equilibrium

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

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Question

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?

Answer

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.

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

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Question

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

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

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

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

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

Answer

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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Question

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?

Answer

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.

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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Question

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

Determine the acid dissociation constant expression for the given reaction.

Answer

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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Question

Calculate the molar solubility of barium fluoride when it is dissolved in a solution containing of sodium fluoride. Barium fluoride has a $\text{K}_{sp}$ value of $2.45\times 10^{-5}$ .

Answer

Start by writing out the equation for the dissolution of barium fluoride.

$\text{BaF}_2(s)\rightleftharpoons\text{Ba}^{2+}(aq)+2\text{F}^-(aq)$

Next, write out the chart to keep track of concentrations of each ion.

| | \[ $Ba^{2+}$ \] | \[\] | | | ------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------ | -------- | | Initial | 0.00 | 0.200 | | Change | +x | +2x | | Equilibrium | x | 0.200+2x |

In the chart, is the amount of barium fluoride that dissolves. Since we are putting barium fluoride in a solution of sodium fluoride, the initial concentration of the fluoride ion is . Since there is already some fluoride in the solution, we should expect that the molar solubility of barium fluoride should be less than its molar solubility when dissolved in water.

Now, write the equilibrium expression using the chemical equation:

$\text{K}_{sp}=[\text{Ba}^{2+}][\text{F}^-]^2$

Plug in the given $\text{K}_{sp}$ and solve for .

$2.45\times10^{-5}=(0.200+2x)^2(x)$

At this point, use a graphing calculator to solve for .

$x=6.05\times10^{-4}$

The molar solubility of barium fluoride in sodium fluoride is $6.05\times10^{-4}M$ .

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Question

Suppose that a chemist wants to lower the concentration of calcium in an aqueous solution by precipitating it out, according to the following reaction expression.

$Ca(OH)_{2}(s)\rightleftharpoons Ca^{2+}(aq)+2OH^{-}(aq)$

Which of the following chemicals could the chemist add to the solution in order to precipitate the calcium?

Answer

In this question, we're shown an equilibrium expression for a precipitation reaction in solution. We're asked to identify a compound that will help calcium precipitate out of solution.

To identify a compound that will precipitate calcium out of solution, we need to consider the equilibrium expression shown in the question stem. Applying Le Chatlier's principle, we'll need to consider the direction each of the answer choices will push the reaction in.

Adding will make the solution more acidic. As a result, there will be a decreased amount of $OH^{-}$ . Consequently, the reaction will shift to the right and there will be more calcium in solution.

Adding $CaCl_{2}$ will drive up the concentration of calcium, which will drive the reaction to the left. Even though this will cause calcium to precipitate out of the solution, it nonetheless defeats the purpose because the amount of dissolved calcium in solution is not decreasing.

Adding to the solution will have no effect on the reaction equilibrium because neither of the ions appears in the expression.

Adding will cause the solution to become more basic, with an increased amount of $OH^{-}$ . Consequently, the reaction equilibrium will be driven toward the left. This will, in turn, cause calcium to precipitate out of solution in the form of the solid $Ca(OH)_{2}$ .

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Question

Consider the following equation at equilibrium:

$2\text{SO}_2(g)+\text{O}_2(g)\rightleftharpoons 2\text{SO}_3(g)$

What would happen if more $\text{O}_2$ were added into the equation?

Answer

Recall Le Chatelier's principle: A chemical system at equilibrium will shift in the direction that minimizes the disturbance to the system.

Before the addition of $\text{O}_2$ , the system is in equilibrium, meaning the reaction quotient is equal to the equilibrium constant; $\text{Q=K}$ .

However, after the addition of $\text{O}_2$ , a reactant, the reaction quotient is now less than the equilibrium constant; $\text{Q}<\text{K}$ .

In order to maintain equilibrium, the reaction will then shift right of favor the formation of more products.

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Question

Consider the following equation at equilibrium:

$4\text{NH}_3(g)+5\text{O}_2(g)\rightleftharpoons4\text{NO}(g)+6\text{H}_2\text{O}(g)$

What would happen if more $\text{NO}$ were added to the system?

Answer

Recall Le Chatelier's principle: A chemical system at equilibrium will shift in the direction that minimizes the disturbance to the system.

Before the addition of $\text{NO}$ , the system is in equilibrium, meaning the reaction quotient is equal to the equilibrium constant; $\text{Q=K}$ .

However, after the addition of $\text{NO}$ , a product, the reaction quotient is now less than the equilibrium constant; $\text{Q}>\text{K}$ .

In order to maintain equilibrium, the reaction will then shift left to favor the reactants in this chemical system.

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Question

Consider the following exothermic equation:

$2\text{C}_8\text{H}_{18}+25\text{O}_2\rightleftharpoons 16\text{CO}_2+18\text{H}_2\text{O}$

Which of the following actions will cause the products to be favored?

Answer

Since the equation is exothermic, you can think of heat as another product of the reaction:

$2\text{C}_8\text{H}_{18}+25\text{O}_2\rightleftharpoons 16\text{CO}_2+18\text{H}_2\text{O}+\text{heat}$

Apply LeChatelier's principle to this equation.

If the temperature at which the reaction is decreased, that is akin to decreasing the amount of product made, thus causing the reaction to shift towards the products.

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Question

$CaF_{2}(s)\rightleftharpoons\ Ca^{2+}(aq)\ +\ 2F^{-}(aq)$

Express the solubility product constant expression for the given reaction.

Answer

The equilibrium given tells us how the solid dissolves in solution:

$CaF_{2}(s)\rightleftharpoons\ Ca^{2+}(aq)\ +\ 2F^{-}(aq)$

However, the solubility product constant (Ksp) given tells us the degree by which a solid dissolves in solution. The larger the Ksp, the more soluble a substance is in water. Writing this expression follows the same rules as other equilibrium constant expressions. Therefore solids and water (when it is the solvent) are omitted from this expression. You must raise the concentration of the substances involved to the power of its coefficient.

For the chemical reaction given, the Ksp is:

$K_{sp}=[Ca^{2+}][F^{-}]^{2}$

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Question

$PbCl_{2}(s)\rightleftharpoons\ Pb^{2+}(aq)\ +\ 2Cl^{-}(aq)$

Express the solubility product constant expression for the given reaction.

Answer

The equilibrium given tells us how the solid dissolves in solution:

$PbCl_{2}(s)\rightleftharpoons\ Pb^{2+}(aq)\ +\ 2Cl^{-}(aq)$

However, the solubility product constant (Ksp) given tells us the degree by which a solid dissolves in solution. The larger the Ksp, the more soluble a substance is in water. Writing this expression follows the same rules as other equilibrium constant expressions. Therefore solids and water (when it is the solvent) are omitted from this expression. You must raise the concentration of the substances involved to the power of its coefficient.

For the chemical reaction given, the Ksp is:

$K_{sp}=[Pb^{2+}][Cl^{-}]^{2}$

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Question

How many grams of $\textrm{BaF}_{\textrm{2}}$ are dissolved in 75.0 mL of a saturated solution of $\textrm{BaF}_{\textrm{2}}$ ?

$\textrm{BaF}_{\textrm{2}}(s)\rightleftharpoons\textrm{Ba}^{\textrm{2+}}(aq)\textrm{ + 2 F}^{\textrm{ -}}(aq)$

$\textrm{K}_{\textrm{sp}}=1.00\textrm{ x 10}^{-6}$

Answer

To calculate how many grams of $\textrm{BaF}_{\textrm{2}}$ are dissolved in the solution, we first solve for the molarity of the solution.

Using the dissociation equation

$\textrm{BaF}_{\textrm{2}}(s)\rightleftharpoons\textrm{Ba}^{\textrm{2+}}(aq)\textrm{ + 2 F}^{\textrm{ -}}(aq)$

We can write out the equation for the solubility product constant as

$\textrm{K}_{\textrm{sp}}=\textrm{[Ba}^{2+}][\textrm{F}^{\textrm{ -}}]^{2}$

$1.00\textrm{ x 10}^{-6}=\textrm{[Ba}^{2+}][\textrm{F}^{\textrm{ -}}]^{2}$

Because there are 2 fluoride ions for every barium ion, we can rewrite the equation as

$[\textrm{Ba}^{2+}]=x$

$1.00\textrm{ x 10}^{-6}=(x)(2x)^{2}$

Now solve for x

$1.00\textrm{ x 10}^{-6}=(x)(4x^{2})$

$1.00\textrm{ x 10}^{-6}=4x^{3}$

$x^{3}=\frac{1.00\textrm{ x 10}^{-6}}{4}=2.50\textrm{ x 10}^{-7}$

$x=\sqrt[3]{2.50\textrm{ x 10}^{-7}}=6.30\textrm{ x 10}^{-3}$

$[\textrm{Ba}^{2+}]=6.30\textrm{ x 10}^{-3}M$

Solve for the concentration of dissolved $\textrm{BaF}_{\textrm{2}}$

$[\textrm{BaF}_{2}]=[\textrm{Ba}^{2+}]=6.30\textrm{ x 10}^{-3}M$

Now that we have the concentration of dissolved $\textrm{BaF}_{\textrm{2}}$ , we can calculate how many grams are dissolved in the solution

$(6.30\textrm{ x 10}^{-3}M)(0.0750\textrm{ L})=4.73\textrm{ x 10}^{-4}\textrm{ mol}$

$(4.73\textrm{ x 10}^{-4}\textrm{ mol})(175.34\frac{\textrm{g}}{\textrm{mol}})=0.0829\textrm{ g BaF}_{2}$

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