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7: Equilibria, Equilibrium Constants and Acid-Bases (Worksheet)

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    79226
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    Name: ______________________________

    Section: _____________________________

    Student ID#:__________________________

    Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

    Learning Objectives

    • Understand the concept of the reaction quotient, \(Q\), as a means of determining whether or not a system is at equilibrium, and if not, how a system must proceed to reach equilibrium
    • Understand how to set up a calculation of all species concentrations at equilibrium, given initial concentrations and the value of the equilibrium constant
    • Understand how various stresses cause a shift in the position of equilibrium on the basis of Le Chatelier’s Principle
    • Understand the Brønsted-Lowry theory of acids and bases
    • Understand the concepts of conjugate acid-base pairs

    When a reaction reaches equilibrium we can calculate the concentrations of all species, both reactants and products, by using information about starting concentrations or pressures and the numerical value of the equilibrium constant. Knowing how to set up and solve equilibrium problems for gas-phase systems is essential preparation for applying equilibrium concepts to more complicated systems, such as acid-base chemistry. The mixture of reactants and products can often be altered by applying a stress to the system (changing species concentrations, changing pressures, changing temperature, etc.), and the shift in the position of the equilibrium can be understood and predicted on the basis of Le Chatelier’s Principle.

    Some the more important applications of equilibrium concepts are concerned with solutions of acids or bases. This week we will look at the definitions of acids and bases in the Brønsted-Lowry theory, and next week we will take up equilibrium calculations of acid-base systems based on that theory.

    Success Criteria

    • Be able to set up and solve for all species using \(K_c\) or \(K_p\)
    • Be able to predict the direction of a reaction on the basis of \(Q\)
    • Be able to check equilibrium calculation results, using a \(Q\) calculation
    • Be able to apply Le Chatelier’s Principle to determine the direction a system at equilibrium must shift to reach a new equilibrium
    • Be able to calculate concentrations once equilibrium is reestablished after a stress that causes a shift from an original equilibrium
    • Be able to analyze acid-base reactions in terms of conjugate acid-base pairs and proton transfer
    • Be able to write the formula of an acid’s conjugate base or a base’s conjugate acid

    The Reaction Quotient: Q

    We can calculate the ratio of concentrations or pressures of reactants and products, like \(K_c\) and \(K_p\), at any time in the course of a reaction, whether or not the system has achieved equilibrium. When the system is not at equilibrium, the ratio of the product concentrations raised to their stoichiometric coefficients to the reactant concentrations raised to their stoichiometric coefficients is called the reaction quotient and given the symbol \(Q\). The form of \(Q\) is the same as the form of \(K_c\) or \(K_p\), but the values for products and reactants are not presumed to be the values at equilibrium. The value of \(Q\) relative to \(K_c\) or \(K_p\) indicates the direction in which the reaction must run to achieve equilibrium. If \(Q < K\), then the reactant concentrations or pressures are too high and the product concentrations or pressures are too low, relative to what they would be at equilibrium. To achieve equilibrium, the reaction must run in the forward direction (shift right), using up reactants and forming more products. Conversely, if \(Q > K\), then the reactant concentrations are too low and the product concentrations are too high, relative to what they would be at equilibrium. To achieve equilibrium, the reaction must run in the reverse direction (shift left), using up products and reforming more reactants. If \(Q = K\), then the system is already at equilibrium.

    Q1

    For the reaction

    \[\ce{H2(g) + I2(g) <=> 2 HI(g)} \nonumber \]

    with \(K_c = 54.8\) at 425 oC. Are the following mixtures of \(\ce{H2}\), \(\ce{I2}\), and \(\ce{HI}\) at 425 oC at equilibrium? If not, in which direction must the reaction proceed to achieve equilibrium?

    1. \(\ce{[H2] = [I2]} = 0.360\, mol/L\), \(\ce{[HI]} = 2.50\, mol/L\)
    2. \(\ce{[H2]} = 0.120\, mol/L\), \(\ce{[I2]} = 0.560\, mol/L\), \(\ce{[HI]} = 2.20 \,mol/L\)
    3. \(\ce{[H2]} = 0.120 \,mol/L\), \(\ce{[I2]} = 0.219\, mol/L\), \(\ce{[HI]} = 1.20\, mol/L\)

    Calculating Amounts of All Species in an Equilibrium Mixture

    Very often we know the initial concentrations or pressures of reactants and products, and we want to know their values when equilibrium is established. To set up a calculation of these amounts, we generally let the variable \(x\) represent the change in an amount of a particular reactant or product that must occur to reach equilibrium. Then, using the stoichiometry of the reaction, we write algebraic expressions to represent the amounts that will be present at equilibrium. We then substitute these algebraic expressions into the concentration or pressure terms of the equilibrium constant expression. Solving for the value of x allows calculating the numerical values of concentration or pressure for all reactants and products. In setting up the problem, it is useful to write down under each species in the balanced equation the initial amount, the algebraic expression for how each amount will change, and the algebraic expressions for each final amount (an ICE table). It is also useful to know the direction in which the reaction must run in order to reach equilibrium. If only reactants are initially present, the reaction must run to the right, forming products. Similarly, if only products are initially present, the reaction must run to the left, forming reactants.

    But if the initial mixture contains amounts of both reactants and products, the direction in which the reaction must run to achieve equilibrium may not be obvious. In such cases, a \(Q\) calculation can be done using the initial concentrations of all species. Remember, if \(Q < K\) the reaction will run in the forward direction, but if \(Q > K\) it will run in the reverse direction. Knowing this at the beginning helps in setting up the algebraic expressions for the changes that must take place to reach equilibrium.

    For example, consider the equilibrium,

    \[\ce{I2(g) + Br2(g) <=> 2 IBr(g)} \nonumber \]

    for which the equilibrium constant \(K_c = 280\) at 150 oC. Suppose 0.500 mol of \(\ce{I2}\) and 0.500 mol of \(\ce{Br2}\) were placed in a one-liter vessel at 150 °C. What will the concentrations of all species be once equilibrium is established?

    Because we have no \(\ce{IBr}\) initially present, the reaction must go to the right in order to reach equilibrium. To write algebraic expressions for the equilibrium concentrations, we could let x equal the concentration of \(\ce{I2}\) that is consumed in forming the equilibrium concentration of \(\ce{IBr}\). But for every mole of \(\ce{I2}\) that reacts, a mole of \(\ce{Br2}\) must also react, so \(x\) also represents the amount of \(\ce{Br2}\) consumed. Because their initial concentrations we each 0.500 mol/L, at equilibrium

    \[ \ce{[I2] = [Br2] = } 0.500 - x, \label{Eq100} \]

    in units of mol/L. At the same time, for every mole of \(\ce{I2}\) and \(\ce{Br2}\) that are consumed, two moles of \(\ce{IBr}\) appear as product, so its concentration at equilibrium will increase by \(2x\). Because there was no \(\ce{IBr}\) initially present, its equilibrium concentration will be \(0 + 2x\); i.e, \(\ce{[Ibr]} = 2x\). From these considerations, our ICE table looks like the following:

    ICE Table \(\ce{I2(g) }\) \(\ce{Br2(g) }\) \(\ce{IBr(g)}\)
    Initial \(0.500\) \(0.500\) \(0\)
    Change \(-x\) \(-x\) \(+2x\)
    Equilibrium \(0.500 - x\) \(0.500 - x\) \(2x\)

    Substituting these algebraic expression into the \(K_c\) expression, we obtain

    \[ \begin{align*} K_c &= \ce{\dfrac{[IBr]^2}{[I2][Br2]}} \\[4pt] &= 280 \\[4pt] &= \dfrac{(2x)^2}{(0.500 - x)(0.500 - x)} \\[4pt] &= \dfrac{4x^2}{(0.500 - x)^2} \end{align*} \]

    We could expand the denominator and rearrange the resulting expression to solve it for \(x\) as a quadratic equation. (Only one root will make sense for the given problem, and the other will be rejected.) However, in this case, the numerator and denominator are both perfect squares, so it would be faster to take the square root of both sides of the expression and then solve for \(x\). Taking the square root of both sides we obtain ,

    \[\sqrt{ 280} = 16.7 = \dfrac{2x}{0.500 -x} \nonumber \]

    Rearranging

    \[ \begin{align*} 8.36 - 16.7 x & = 2x \\[4pt] 18.7 x &=8.36 \\[4pt] x &= 0.446 \label{X} \end{align*} \nonumber \]

    We can now substitute \(x\) back into our algebraic expression to obtain numerical values for each of the concentrations (Equation \ref{Eq100}).

    \[ [\ce{I2}] = [\ce{Br2}] = 0.500 – x = 0.500 – 0.446 = 0.053 \,mol/L \nonumber \]

    \[[\ce{IBr}] = 2x = (2)(0.446 ) = 0.893 \,mol/L \nonumber \]

    Q2

    For the reaction

    \[ \ce{(NH3)B(CH3)3 (g) <=> NH3 (g) + B(CH3)3 (g)} \nonumber \]

    at 100 °C and \(K_p = 4.62 \,atm\). If the partial pressures of \(\ce{NH3(g)}\) and \(\ce{B(CH3)3 (g)}\) in an equilibrium mixture at 100 °C are both 1.52 atm, what is the partial pressure of \(\ce{(NH3)B(CH3)3 (g)}\) in the mixture?

    Q3

    At 425 °C, 1.00 mol of \(\ce{H2(g)}\) and 1.00 mol of \(\ce{I2(g)}\) are mixed in a one liter vessel. What will be the concentrations of \(\ce{H2 (g)}\), \(\ce{I2(g)}\), and \(\ce{HI(g)}\) at equilibrium. K = 54.8 at 425 °C for the reaction:

    \[\ce{H2 (g) + I2 (g) <=> 2 HI(g)} \nonumber \]

    Using Q to Check Your Results

    In general, equilibrium calculations involve a lot of mathematical manipulation, and it is easy to make a mistake. How can you know when you have gone astray? The best way is to substitute your found values into the expression for \(Q\) and compare the calculated value with the given \(K\). Because of rounding, do not expect an exact match. However, if \(Q \neq K\) by a wide margin, check your work. In our example calculation of the equilibrium concentrations of \(\ce{I2(g)}\), \(\ce{Br2(g)}\), and \(\ce{IBr(g)}\), a \(Q\) calculation with the values we found gives the stated value of \(K_c\).

    \[ \begin{align*} Q_c & = \ce{\dfrac{[IBr]^2}{[I2][Br2]}} \\[4pt] &= \dfrac{0.893}{(0.053)(0.053)} \\[4pt] &= 280 = K_c \end{align*} \nonumber \]

    Thus, we have confidence that the calculation is correct. When doing this kind of check, realize that arithmetic loss of significant figures and rounding differences may cause the calculated value of \(Q\) to be somewhat different from \(K\), but the difference should not be great.

    Q4

    Check the values you found in Q3 by calculating the value of \(Q\). Does your value c agree with \(K_c = 54.8\)?

    Q5

    Suppose \(0.800\, mol\, \ce{H2(g)}\), \(0.900 \,mol \, \ce{I2 (g)}\), and \(0.100\, mol\, \ce{HI(g)}\) are mixed in a one liter vessel at 425 °C. \(K_c = 54.8\) at 425 °C for the reaction

    \[\ce{H2(g) + I2(g) <=> 2HI(g)} \nonumber \]

    1. In which direction must the reaction run (forward or backwards) to achieve equilibrium?
    2. What are the concentrations of all species at equilibrium? Check your final answers with a \(Q\) calculation.

    Le Chatelier's Principle

    In 1884 Henri Le Chatelier formulated the following principle:

    If a stress is applied to a system at equilibrium, the system will tend to adjust to a new equilibrium, which minimizes the stress, if possible.

    The stresses are changes in concentration, pressure, and temperature. If the stress causes a change in the amounts of reactants and products present once equilibrium is reestablished, we say that a shift in the position of the equilibrium has occurred. If we think of a reaction equation in the usual way, with reactants on the left and products on the right, a stress to the system at equilibrium may cause either of the following:

    • Shift to the right: more reactant(s) consumed resulting in greater product and lesser reactant concentrations
    • Shift to the left: more product(s) consumed resulting in greater reactant and lesser product concentrations

    Sometimes the stress cannot be alleviated by either kind of shift, in which case the original equilibrium is maintained. The effects of each kind of stress on a system at equilibrium are summarized below.

    Concentration change

    • \(K_c\) or \(K_p\) remains the same.
    • Increasing reactant concentrations or decreasing product concentrations causes a shift right (more product forms).
    • Increasing product concentrations or decreasing reactant concentrations causes a shift left (more reactant forms).

    Pressure change

    • \(K_c\) or \(K_p\) remains the same.
    • Only changes that affect the partial pressures of reactants and/or products can cause a change. For example, adding an inert gas has no effect.
    • Increasing pressure causes a shift to the side with the lower sum of coefficients on gas species. For example, increasing the pressure on an equilibrium mixture for the reaction, \[\ce{N2 (g) + 3 H2 (g) <=> 2 NH3 (g)}, \nonumber \] causes a shift right.
    • If the sum of coefficients on gas species is the same on the left and right, changing the pressure has no effect. For example, for an equilibrium mixture for the reaction, \[\ce{H2 (g) + I2 (g) <=> 2 HI(g)}, \nonumber \] changing the pressure has no effect on the position of the equilibrium.

    Temperature change

    • \(K_c\) and \(K_p\) values change!
    • Raising the temperature drives the endothermic process. For example, for the reaction, \[\ce{N2O4 (g) <=> 2 NO (g)} \nonumber \] with \(\Delta H = +58.0\, kJ/mol\). Therefore, raising the temperature on an equilibrium mixture will favor formation of more \(\ce{NO2 (g)}\) because the forward reaction is endothermic.
    • Lowering the temperature drives the exothermic process. For example, for the reaction, \[\ce{N2O4 (g) <=> 2 NO (g)} \nonumber \] with \(\Delta H = +58.0\, kJ/mol\). Therefore, lowering the temperature on an equilibrium mixture will favor formation of more \(\ce{N2O4 (g)}\), because the reverse reaction is exothermic.

    Adding a catalyst

    • A catalyst has no effect on the position of the equilibrium, just how fast it gets there.

    Q6

    For each of the following reactions at equilibrium, predict the effect (if any) the indicated stress would have on the position of the equilibrium. Note whether or not the value of the equilibrium constant changes.

    1. \(\ce{H2 (g) + I2 (g) <=> 2 HI(g)}\) with more \(\ce{HI(g)}\) added.
    2. \(\ce{N2 (g) + 3 H2 (g) <=> 2 NH3 (g)}\) with \(\ce{NH3(g)}\) is removed as it forms.
    3. \(\ce{2 NO(g) + Cl2 (g) <=> 2 NOCl(g)}\) with overall pressure is increased.
    4. \(\ce{2 NO(g) + Cl2 (g) <=> 2 NOCl(g)}\) \(\Delta H = –75.5\, kJ/mol\) and temperature is increased.
    5. \(\ce{H2O(g) + C(s) <=> H2(g) + CO(g)}\) with overall pressure is increased.
    6. \(\ce{C(s) + O2(g) <=> CO2 (g)}\) with overall pressure is decreased.
    7. \(\ce{N2 O4 (g) <=> 2 NO2 (g)}\) with \(\ce{N2 (g)}\) is added, increasing overall pressure.
    8. \(\ce{N2(g) + 3 H2 (g) <=> 2 NH3 (g)}\) with iron powder added as a catalyst.

    Q7

    An equilibrium mixture of \(\ce{H2(g)}\), \(\ce{I2 (g)}\), and \(\ce{HI(g)}\) in a one-liter vessel at 425 °C is found to have the following concentrations:

    • \(\ce{[H2 ]} = 0.146\, mol/L\)
    • \(\ce{[I2 ]} = 0.246\, mol/L\)
    • \(\ce{[HI]} = 1.41\, mol/L\)

    If \(0.59\, mol\) of \(\ce{HI}\) is added to the vessel, what will be the concentrations of all species once equilibrium is reestablished? \(K = 54.8\) at 425 °C for the reaction:

    \[\ce{H2 (g) + I2(g) <=> 2 HI(g)} \nonumber \]

    Brønsted-Lowry Acid-Base Theory

    In the Arrhenius theory, an acid is a substance that produces \(\ce{H^{+}}\) ion in solution, and a base is a substance that produces \(\ce{OH^{-}}\) ion in solution. This theory is only useful for aqueous (water) – solutions. In 1923, Brønsted (Danish) and Lowry (English) independently proposed a new theory that could be applied to other solvent systems. The Brønsted-Lowry Theory uses the following definitions:

    • Acid - a substance that donates protons (\(\ce{H^{+}}\))
    • Base - a substance that accepts protons (\(\ce{H^{+}}\))

    In this theory, acid-base reactions are seen as proton transfer reactions. For example, consider the reaction between the hydronium ion and ammonia in water:

    \[\ce{H3O^{+} (aq) + NH3 (aq) <=> H2O(l) + NH4^{+}(aq) } \nonumber \]

    This involves transfer of a proton from the hydronium ion to the ammonia molecule to produce water and the ammonium ion. This make \(\ce{H3O^{+}}\) an acid and \(\ce{NH3}\) a base. In general terms, all Brønsted-Lowry acid-base reactions fit the general pattern

    \[\underbrace{\ce{HA}}_{\text{acid}} + \underbrace{\ce{B}}_{\text{base}} \ce{ <=> A^{-} + HB^{+}} \label{EQ1}\]

    acid base When an acid, HA, loses a proton it becomes its conjugate base, \(\ce{A^{-}}\), a species capable of accepting a proton in the reverse reaction.

    \[\underbrace{\ce{HA}}_{\text{acid}} \ce{ <=>} \underbrace{\ce{A^{-}}}_{\text{Conjugate} \\ \text{Base}} + \ce{H^{+}} \label{EQ2}\]

    Likewise, when a base, \(\ce{B}\), gains a proton, it becomes its conjugate acid, \(\ce{BH}\), a species capable of donating a proton in the reverse reaction.

    \[\underbrace{\ce{B}}_{\text{base}} + \ce{H^{+}} \ce{<=>} \underbrace{\ce{HB^{+}}}_{\text{Conjugate} \\ \text{Acid}} \nonumber \]

    The Brønsted-Lowry concept of conjugate acid base pairs leads to the idea that all acid-base reactions are proton transfer reactions. The generic reaction between an acid \(\ce{HA}\) and a base \(\ce{B}\) can be viewed as a sum of two reactions (Equation \ref{EQ1} and \ref{EQ2}):

    \[ \begin{align*} \ce{HA \, & <=>\, A^{-}} + \cancel{\ce{H^{+}}} \\ \cancel{\ce{ H^{+}}} + \ce{ B \, &<=> \, HB^{+}} \end{align*} \]

    \[\underbrace{\ce{HA}}_{\text{acid}_1} + \underbrace{B}_{\text{base}_2} \ce{<=> } \underbrace{A}_{\text{base}_1} + \underbrace{ \ce{HB^{+}} }_{\text{acid}_2} \nonumber \]

    Species with the same subscripts are a conjugate acid-base pairs.

    Q8

    In the spaces provided, write the formulas of the conjugate bases of the given Brønsted-Lowry acids. Remember to write the proper charge, if any, for the conjugate base in each case.

    Acid \(\ce{HNO3}\) \(\ce{NH4^{+} }\) \(\ce{HSO4^{-} }\) \(\ce{H2SO4 }\) \(\ce{HC2H3O2 }\)
    Conjugate base

    Q9

    In the spaces provided, write the formulas of the conjugate acids of the given Brønsted-Lowry bases. Remember to write the proper charge, if any, for the conjugate acid in each case.

    Base \(\ce{OH^{-}}\) \(\ce{NH3}\) \(\ce{HS^{-} }\) \(\ce{SO4^{2-}}\) \(\ce{CH3NH2}\)
    Conjugate Acid

    Q10

    Using the notation acid /base and base /acid (as shown above), identify the Brønsted-Lowry conjugate acid-base pairs in the following reactions.

    1. \(\ce{HS^{-} (aq) + HC2H3O2 (aq) <=> H2S + C2H3O2^{2-} }\)
    2. \(\ce{HF (aq) + PO4^{3-} (aq) <=> F^{-} (aq) + HPO4^{2-} (aq) }\)
    3. \(\ce{HNO2 (aq) + H2O(l) <=> NO2^{-} (aq) + H3O^{+}}\)
    4. \(\ce{CH3NH2 (aq) + H2O(l) <=> CH3NH3^{+} (aq) + OH^{-} (aq)}\)

    7: Equilibria, Equilibrium Constants and Acid-Bases (Worksheet) is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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