25.14: Questions
- Page ID
- 32794
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Quick questions
You should be able to answer these questions without too much difficulty after studying this TLP. If not, then you should go through it again!
The Ellingham diagram shows values of which thermodynamic quantity as a function of temperature?
- Answer
-
B
For a closed system at equilibrium at a temperature T, which of the following statements are true?
- ΔG = 0;
- ΔG° = 0;
- ΔH = TΔS;
- ΔS = 0.
- Answer
-
a. is true; and hence c. is true. At equilibrium, there is no driving force for a reaction to proceed in any direction - no decrease in free energy occurs. Therefore, ΔG = ΔH - TΔS = 0. The enthalpy change of the reaction is balanced by the entropy change of the reaction. ΔG° is the standard free energy change for the reaction, which is the free energy change when all the reactants and products are in their standard states.
Why are the slopes of many of the lines on the Ellingham diagram almost identical?
- Answer
-
B. Most of the reactions on the Ellingham diagram involve the elimination of one mole of gas to form a solid product. Since the entropy of a gaseous state is very much greater than a solid state, but the entropy of different gases is similar, most reactions have a similar entropy change.
What thermodynamic quantity does the intercept at T=0 K for any standard free energy vs T line signify?
- Answer
-
B. Since ΔG = ΔH - TΔS, at T =0, ΔG = ΔH
What is the decomposition temperature (to the nearest 50 K) for Ag2O? (This question should be completed with the help of the interactive Ellingham diagram included with this TLP).
- Answer
-
A
What is the decomposition temperature (to the nearest 50 K) for PdO? (This question should be completed with the help of the interactive Ellingham diagram included with this TLP).
Hint: On the Ellingham diagram, select the 'oxides' reactions, and then select the elements Pd from the periodic table. The decomposition temperature is the temperature at which the standard free energy of formation of the oxide becomes positive. ie. ΔG° = 0.
- Answer
-
C
Which of the following elements can be used to produce Cr from Cr2O3 at 1200K? (This question should be completed with the help of the interactive Ellingham diagram included with this TLP).
- Mg
- Fe
- Co
- Al
Hint: Display the oxide reactions for chromium along with each of the other elements on the Ellingham diagram. What does the relative stability of the oxides of each element and Cr2O3 (shown by their relative position on the diagram) tell you?
- Answer
-
a and d. The standard free energy change for the formation of MgO and Al2O3 is more negative than for Cr2O3. This means that either of these compounds will reduce Cr2O3 to Cr.
Explain, using the Ellingham diagram, how a mixture of Cl2/O2 gas may be used to separate Zn from Fe in a galvanised scrap. (This question should be completed with the help of the interactive Ellingham diagram included with this TLP).
- Answer
-
On the Ellingham diagram, the free energy of formation for iron oxide is always lower than the free energy of formation of iron chloride. Hence, iron will always react to form its oxide in a Cl2/O2 gas mixture. Conversely, the free energy of formation of zinc oxide is always greater than the free energy of formation of zinc chloride, so zinc will always form its chloride in a Cl2/O2 gas mixture.
In addition, any iron chloride will be reduced by zinc metal, and any zinc oxide will be reduced by iron.