Chemical Equilibrium (McGuire)
- Page ID
- 281945
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)1.) Write the equilibrium constant expression for each of the following reactions.
- \(\ce{2 H2O2 (g) ⇌ 2H2O (g) + O2 (g)}\)
- \(\ce{6 H2O2 (g) ⇌ 6 H2O (g) + 3 O2 (g)}\)
- The reverse of the reaction in part a
- \(\ce{2PbS (s) + 3O2 (g) ⇌ 2PbO (s) + 2SO2 (g)}\)
- \(\ce{MgCl2(s) ⇌ Mg^2+ (aq) + 2 Cl- (aq)}\)
- The reverse of the reaction in part e
2.) Calculate the equilibrium constant for this reaction:
\[\ce{2 PO2Br (aq) ⇆ 2 PO2 (aq) + Br2 (aq)}\nonumber\]
Given: [PO2Br] = 0.0255M, [PO2] = 0.155M, and [Br2] = 0.00351M at equilibrium.
3.) A solution is prepared having the following initial concentrations:
\[\ce{[Fe^3+ ]} = \ce{[Hg2^2+ ]} = \textrm{0.5000 M;} \hspace{30px} \ce{[Fe^2+ ]} = \ce{[Hg^2+ ]} = \textrm{0.03000 M}\nonumber\]
The following reaction occurs among the ions at a certain temperature.
\[\ce{2Fe^3+(aq) + Hg2^2+(aq) ⇔ 2Fe^2+(aq) + 2Hg^2+(aq)} \hspace{30px} \mathrm{K = 9.14 \times 10^{-8}}\nonumber\]
Will the concentration of each the ions be higher or lower when equilibrium is established?
4.) An important exothermic reaction in the commercial production of hydrogen is
\[\ce{CO(g) + H2O(g) ⇔ H2(g) CO2(g)}\nonumber\]
How will the system at equilibrium shift in each of the following cases?
- CO2 is removed.
- H2O(g) is added.
- The pressure is increased by adding helium gas.
- The temperature is increased.
- The pressure is increased by decreasing volume.
5.) Consider the endothermic reaction
\[\ce{CaCO3(s) ⇔ CaO(s) + CO2(g)}\nonumber\]
Fill in the table below showing how the indicated change to the equilibrium system will affect the indicated quantity when a new equilibrium state is established.
Change |
Quantity |
Effect |
---|---|---|
Add CaO(s) |
CO2 |
|
Decrease the container volume |
CaO(s) |
|
Decrease the container volume |
Total Pressure |
|
Add CO2 |
K |
|
Add CO2 |
CaCO3 |
|
Add Helium gas |
CO2 |
|
Increase the temperature |
CO2 |
|
Increase the temperature |
K |
|
6.) At 1000 K, the value of K for the reaction 2SO3(g) = 2SO2(g) + O2(g) is 0.338. Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial pressures of reactants are PSO3 = 2 x 10-3 atm; PSO2 = 5 x 10-3 atm; PO2 = 3 x 10-2 atm.
7.) Consider this endothermic reaction: 3 O2(g) ⇆ 2 O3(g). To shift this reaction towards the reactants:
- You could ___________ the pressure.
- You could ___________ the volume.
- You could ___________ oxygen gas.
- You could ___________ the temperature.
8.) Which of the following, if increased, will change the value of the equilibrium constant in the example above?
- Pressure b. Volume c. [Product] d. Temperature e. [Reactant]
9.) For the following equilibrium
\[\ce{CH3COOH ⇔ CH3COO- + H+}\nonumber\]
If K=1.8x10-5, at what [H+] concentration will there be equal concentrations of CH3COOH and CH3COO-?
10.) At 298 K, F3SSF (g) decomposes partially to SF2 (g). At equilibrium, the partial pressure of SF2 (g) is 1.1 × 10-4 atm and the partial pressure of F3SSF is 0.0484 atm.
- Write a balanced equilibrium equation to represent this reaction.
- Compute the equilibrium constant corresponding to the equation you wrote.
11.) The equilibrium constant for the reaction
\[\ce{H2S} (\textit{g}) + \ce{I2} (\textit{g}) ⇋ \ce{2 HI} (\textit{g}) + \ce{S} (\textit{s})\nonumber\]
at 110°C is equal to 0.0023. Calculate the reaction quotient Q for each of the following conditions and determine whether solid sulfur is consumed or produced as the reaction comes to equilibrium.
- PI2 = 0.461 atm; PH2S = 0.050 atm; PHI = 0.0 atm
- PI2 = 0.461 atm; PH2S = 0.050 atm; PHI = 9.0 atm
Contributors and Attributions
- Molly McGuire, Bucknell University (mmcguire@bucknell.edu)
- Sourced from the Analytical Sciences Digital Library