19.13: The Temperature Dependence of ΔH
- Page ID
- 13711
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)It is often required to know thermodynamic functions (such as enthalpy) at temperatures other than those available from tabulated data. Fortunately, the conversion to other temperatures is not difficult.
At constant pressure
\[ dH = C_p \,dT \nonumber \]
And so for a temperature change from \(T_1\) to \(T_2\)
\[ \Delta H = \int_{T_2}^{T_2} C_p\, dT \label{EQ1} \]
Equation \ref{EQ1} is often referred to as Kirchhoff's Law. If \(C_p\) is independent of temperature, then
\[\Delta H = C_p \,\Delta T \label{intH} \]
If the temperature dependence of the heat capacity is known, it can be incorporated into the integral in Equation \ref{EQ1}. A common empirical model used to fit heat capacities over broad temperature ranges is
\[C_p(T) = a+ bT + \dfrac{c}{T^2} \label{EQ15} \]
After combining Equations \ref{EQ15} and \ref{EQ1}, the enthalpy change for the temperature change can be found obtained by a simple integration
\[ \Delta H = \int_{T_1}^{T_2} \left(a+ bT + \dfrac{c}{T^2} \right) dT \label{EQ2} \]
Solving the definite integral yields
\[ \begin{align} \Delta H &= \left[ aT + \dfrac{b}{2} T^2 - \dfrac{c}{T} \right]_{T_1}^{T_2} \\ &= a(T_2-T_1) + \dfrac{b}{2}(T_2^2-T_1^2) - c \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \label{ineq} \end{align} \]
This expression can then be used with experimentally determined values of \(a\), \(b\), and \(c\), some of which are shown in the following table.
Substance | a (J mol-1 K-1) | b (J mol-1 K-2) | c (J mol-1 K) |
---|---|---|---|
C(gr) | 16.86 | 4.77 x 10-3 | -8.54 x 105 |
CO2(g) | 44.22 | 8.79 x 10-3 | -8.62 x 105 |
H2O(l) | 75.29 | 0 | 0 |
N2(g) | 28.58 | 3.77 x 10-3 | -5.0 x 104 |
Pb(s) | 22.13 | 1.172 x 10-2 | 9.6 x 104 |
What is the molar enthalpy change for a temperature increase from 273 K to 353 K for Pb(s)?
Solution
The enthalpy change is given by Equation \ref{EQ1} with a temperature dependence \(C_p\) given by Equation \ref{EQ1} using the parameters in Table \(\PageIndex{1}\). This results in the integral form (Equation \ref{ineq}):
\[ \Delta H = a(T_2-T_1) + \dfrac{b}{2}(T_2^2-T_1^2) - c \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \nonumber \]
when substituted with the relevant parameters of Pb(s) from Table \(\PageIndex{1}\).
\[ \begin{align*} \Delta H = \,& (22.14\, \dfrac{J}{mol\,K} ( 353\,K - 273\,K) \\ & + \dfrac{1.172 \times 10^{-2} \frac{J}{mol\,K^2}}{2} \left( (353\,K)^2 - (273\,K)^2 \right) \\ &- 9.6 \times 10^4 \dfrac{J\,K}{mol} \left( \dfrac{1}{(353\,K)} - \dfrac{1}{(273\,K)} \right) \\ \Delta H = \, & 1770.4 \, \dfrac{J}{mol}+ 295.5\, \dfrac{J}{mol}+ 470.5 \, \dfrac{J}{mol} \\ = & 2534.4 \,\dfrac{J}{mol} \end {align*} \]
For chemical reactions, the reaction enthalpy at differing temperatures can be calculated from
\[\Delta H_{rxn}(T_2) = \Delta H_{rxn}(T_1) + \int_{T_1}^{T_2} \Delta C_p \Delta T \nonumber \]
The enthalpy of formation of NH3(g) is -46.11 kJ/mol at 25 oC. Calculate the enthalpy of formation at 100 oC.
Solution
\[\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)} \nonumber \]
with \(\Delta H \,(298\, K) = -46.11\, kJ/mol\)
Compound | Cp (J mol-1 K-1) |
---|---|
N2(g) | 29.12 |
H2(g) | 28.82 |
NH3(g) | 35.06 |
\[ \begin{align*} \Delta H (373\,K) & = \Delta H (298\,K) + \Delta C_p\Delta T \\ & = -46110 +\dfrac{J}{mol} \left[ 2 \left(35.06 \dfrac{J}{mol\,K}\right) - \left(29.12\, \dfrac{J}{mol\,K}\right) - 3\left(28.82\, \dfrac{J}{mol\,K}\right) \right] (373\,K -298\,K) \\ & = -49.5\, \dfrac{kJ}{mol} \end{align*} \]