Solutions
- Page ID
- 90923
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Using the tables on the web, calculate DSo and DGo for the following reactions using:
DSo = SSoproducts - SSoreactants and
DGorxn=SDGoproducts - SDGoreactants
1. CH4 (g) + 2 O2 (g) ---------> CO2 (g) + 2 H2O (l)
a) DSo = SSoproducts - SSoreactants
| Soproducts | Soreactants |
| (1 mol CO2)(213.7 J/Kmol) = 213.7 J/K | (1 mol CH4)(186.1 J/Kmol) = 186.1 J/K |
| (2 mol H2O)(69.940 J/Kmol) = 139.88 J/K | (2 mol O2)(205 J/Kmol) = 410 J/K |
| SSoproducts= 353.58 J/K | SSoreactants = 596.1 J/K |
| DSo= 353.58 J/K - 596.1 J/K = -242.52 J/K |
b) DGorxn=SDGoproducts - SDGoreactants
| DGoproducts | DGoreactants |
| (1 mol CO2)(-394.4 kJ/mol) = -394.4 kJ | (1 mol CH4)(-50.81 kJ/mol) = -50.81 kJ |
| (2 mol H2O)(-237.192 J/Kmol) = -474.384 kJ | (2 mol O2)(0 kJ/mol) = 0 kJ |
| SDGoproducts = -868.784 kJ | SDGoreactants= -50.81 kJ |
| DGorxn= -868.784 kJ - (-50.81 kJ) = -817.974 kJ | |
2. 2 MgO (s) --------> 2 Mg (s) + O2 (g)
a) DSo = SSoproducts - SSoreactants
| Soproducts | Soreactants |
| (2 mol Mg)(32.69 J/Kmol) = 65.38J/K | (2 mol MgO)(26.9 J/Kmol) = 53.8 J/K |
| (1 mol O2)(205 J/Kmol) = 205 J/K | |
| SSoproducts = 270.38 J/K | SSoreactants = 53.8 J/K |
| DSo = 270.38 J/K - 53.8 J/K = 216.58 J/K |
b) DGorxn=SDGoproducts - SDGoreactants
| DGoproducts | DGoreactants |
| (2 mol Mg)( 0 kJ/mol) = 0 kJ | (2 mol MgO)(-569.0 kJ/mol) = -1138.0 kJ |
| (1 mol O2)( 0 kJ/mol) = 0 kJ | |
| SDGoproducts = 0 kJ | SDGoreactants = -1138.0 kJ |
| DGorxn= 0 kJ - (-1138.0 kJ) = +1138.0 kJ | |