9.E: Fundamental Equlibrium Concepts - Homework
- Page ID
- 441049
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Chemical Equilibria
- Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?
- Answer
- yes
- If the concentrations of products and reactants are equal, is the system at equilibrium?
- Answer
- no
Equilibrium Constants
- Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature.
- Answer
- the reaction quotient depends upon the concentration of the products and reactants, which can have an infinate number of values. The equilibrium constant is a constant - so it only has one value
- Explain why an equilibrium between Br2(l) and Br2(g) would not be established if the container were not a closed vessel shown in Figure 7.1.4.
- Answer
- in an open vessel the bromine gas can escape and is not able to condense back into a liquid - so the system can not go in both directions
- If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO2 or with pure N2O4? \( \mathrm{2 NO_2} \rightleftharpoons \mathrm{N_2O_4} \)
- Answer
- no
- Write the mathematical expression for the reaction quotient, Qc, for the following reaction: \( \mathrm{CH_4}(g) + \mathrm{Cl_2}(g) \rightleftharpoons \mathrm{CH_3Cl}(g) + \mathrm{HCl}(g) \)
- Answer
- \[ Q=\frac{[\mathrm{CH_3Cl}][\mathrm{HCl}]}{[\mathrm{CH_4}][\mathrm{Cl_2}]} \]
- Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium? \( \mathrm{SO_2Cl_2}(g) \rightleftharpoons \mathrm{SO_2}(g) + \mathrm{Cl_2}(g) \) ; [SO2Cl2] = 0.12 M, [Cl2] = 0.16 M and [SO2] = 0.050 M. Kc for the reaction is 0.078.
- Answer
- Q=0.66, Q is less than K so the reaction will go forward, increasing the concentration of products and reducing the concentration of reactants.
Shifting Equilibria: Le Châtelier’s Principle
- Methanol can be prepared from carbon monoxide and hydrogen at high temperature and pressure in the presence of a suitable catalyst. \[ \mathrm{H_2}(g) + \mathrm{CO}(g) \rightleftharpoons \mathrm{CH_3OH}(g)\]
- (a) Write the expression for the equilibrium constant (K) for the reversible reaction
- Answer
- \[ Q=\frac{[\mathrm{CH_3OH}]} {[\mathrm{H_2}][\mathrm{CO}]} \]
- (b) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more H2 is added?
- Answer
- H2 and CO will decrease, CH3OH will increase
- (c) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CO is removed?
- Answer
- H2 and CO will increase, CH3OH will decrease
- (d) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if CH3OH is added?
- Answer
- H2 and CO will increase, CH3OH will decrease
- (e) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if the temperature of the system is increased (Δ H = -90.2 kJ)?
- Answer
- H2 and CO will increase, CH3OH will decrease
- (f) What will happen to the concentrations of H2, CO, and CH3OH at equilibrium if more catalyst is added?
- Answer
- no change
- (a) Write the expression for the equilibrium constant (K) for the reversible reaction
- Water gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon. \[ \mathrm{C}(s) + \mathrm{H_2O}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H_2}(g) \]
- (a) Write the expression for the equilibrium constant for the reversible reaction
- Answer
- \[ Q=\frac{[\mathrm{CO}][\mathrm{H_2}]} {[\mathrm{H_2O}]} \]
- (b) What will happen to the concentration of each reactant and product at equilibrium if more C is added?
- Answer
- no change, since C is a solid and is not in the equlibrium constant expression
- (c) What will happen to the concentration of each reactant and product at equilibrium if H2O is removed?
- Answer
- amount of C will increase, concentration of H2O will increase, CO and H2 will decrease
- (d) What will happen to the concentration of each reactant and product at equilibrium if CO is added?
- Answer
- amount of C will increase, concentration of H2O will increase, CO and H2 will decrease
- (e) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased? Δ H = 131.30 kJ
- Answer
- amount of C will decrease, concentration of H2O will decrease, CO and H2 will increase
- (a) Write the expression for the equilibrium constant for the reversible reaction


