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Suppose we have an arbitrary chemical reaction $A+B\rightleftharpoons 2C+D$ and its equilibrium constant at two temperatures $T_{1},T_{2}$ are $k_{1},k_{2}$. We can relate them as $$\log\frac{k_{2}}{k_{1}}=\frac{\Delta H}{2.303R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)\ \ \ \ -(i)$$ I am unclear on the significance of $\Delta H$ in the above equation.

Does $\Delta H$ represent the enthalpy change of the reaction at any given temperature or is it the difference of reaction enthalpies at the two different temperatures (given by $\Delta H_{2}-\Delta H_{1}=C_{p,reaction}\Delta T$).

I tried to go back to the Arrhenius equation to understand this. Arrhenius' equation gives a relation for the forward rate constant $k_{f}$ of the reaction with its activation energy $E_{a}$ $$k_{f}=Ae^{-\frac{E_{a}}{RT}}$$ and we take $k_{eq}=\frac{k_{f}}{k_{b}}$. However, it doesn't improve my understanding on the subject because we have to assume $\Delta H$ to be constant with temperature to get equation (i).

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  • $\begingroup$ There is convention to keep TD constants as K while kinetic constants as k, what comes handy if mixing them up. // See also Van 't Hoff equation $$\frac{d}{dT} \ln K_\mathrm{eq} = \frac{\Delta_r H^\ominus}{RT^2},$$ of which is your equation above the integral form. The equation implies $\Delta H_r$ is constant on the interval $T_1 - T_2$. $\endgroup$
    – Poutnik
    Jun 2, 2023 at 11:37
  • $\begingroup$ The activation energy is that energy needed to reach the transition state (top of the hill as it were ) between products and reactants and is not the same as $\Delta H$ which is the enthalpy change from reactants to products at equilibrium so you cannot mix them or get one from the other. $\endgroup$
    – porphyrin
    Jun 2, 2023 at 11:39
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    $\begingroup$ There is error in you formula, there is just /R, not /(RT) // Easy check will tell you that both equation sides have to be unitless, what is not possible with the extra T. $\endgroup$
    – Poutnik
    Jun 2, 2023 at 12:25

3 Answers 3

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There is no need to invoke the Arrhenius equation to understand it. I will give the various form of that equation so you are out of doubts.

The significance is clear when you go back to the differential form, this is, van't Hoff's equation $$ \boxed{\frac{\mathrm{d}\ln K}{\mathrm{d}T} = \frac{\Delta_\mathrm{r} H^\circ(T)}{RT^2}} \tag{1}$$ in words: $\Delta_\mathrm{r} H^\circ$ is the enthalpy of reaction when $1$ mol of $A$ in its standard state at temperature $T$ reacts with $1$ mol of $B$ in its standard state at temperature $T$, to yield $2$ mol of $C$ in its standard state at temperature $T$ and $1$ mol of $D$ in its standard state at temperature $T$.

The following points are true:

  • The value of $\Delta_\mathrm{r} H^\circ$ depends on temperature.
  • The integration of Eq. (1) assuming its constant value is (there is a typo in your equation) $$ \ln\left[\frac{K(T_2)}{K(T_1)}\right] = -\frac{\Delta_\mathrm{r} H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \tag{2}$$ this integration is only carried on exams or easy exercises on books. However, if you are dealing with a new reaction, it is a nice way to get an estimate of $K(T_2)$. Its validity is simple: the estimation will be worse as the distance from $T_1$ to $T_2$ is large. It is also an excellent guess when the reaction is not that exothermic neither endothermic, so large temperature changes are needed to shake the value of $\Delta_\mathrm{r} H^\circ(T)$.
  • The dependence of $\Delta_\mathrm{r} H^\circ$ can be obtained by doing an additional integration \begin{align} \frac{\mathrm{d}\Delta_\mathrm{r} H^\circ}{\mathrm{d}T} &= \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) \\ \int_{\Delta_\mathrm{r} H^\circ(T_1)}^{\Delta_\mathrm{r} H^\circ(T_2)} \mathrm{d}\Delta_\mathrm{r} H^\circ &= \int_{T_1}^{T_2} \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) \; \mathrm{d}T \\ \Delta_\mathrm{r} H^\circ(T_2) &= \Delta_\mathrm{r} H^\circ(T_1) + \int_{T_1}^{T_2} \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) \; \mathrm{d}T \tag{3} \\ \end{align} where in your reaction \begin{align} \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) &= \sum_\mathrm{j} C_\mathrm{p,j}^\circ(T) \\ \Delta_\mathrm{r} C_\mathrm{p}^\circ(T) &= 2C_\mathrm{p,C}^\circ(T) + C_\mathrm{p,D}^\circ(T) - C_\mathrm{p,A}^\circ(T) - C_\mathrm{p,B}^\circ(T) \tag{4} \end{align} In Eq. (3), typically you have access (at the back of the books) for the value $ \Delta_\mathrm{r} H^\circ(T_1) $ at $T_1 = \pu{298.15 K}$. Thus, if you get heat capacity data for your substances $A$, $B$, $C$, and $D$ as a function of temperature, you can combine Eqs. (3) and (4) and just integrate.
  • You have to respect the heat capacites in Eq. (4). If $A$ is a liquid, then the heat capacity must be that of a real liquid for the pressure of $\pu{1 bar}$. If it is a gas, then the heat capacity must be that of an ideal gas for the pressure of $\pu{1 bar}$.
  • When you do the full integration, but not for a temperature $T_2$ but arbitrary temperature $T$, then you can go back to Eq. (1). Once you integrate it again, you have the "exact" value of $K(T_2)$, in the sense that you get a more exact value the better your data is.
  • The worst assumption is $\Delta_\mathrm{r} H^\circ = \text{constant}$. If you feel strong, you perform the full integration. An intermediate level is to consider the functionality of temperature, but to assume that the heat capacities of $A$, $B$, $C$, and $D$ are constant in $T \in [T_1,T_2$]. Eq. (3) becomes $$ \Delta_\mathrm{r} H^\circ(T_2) = \Delta_\mathrm{r} H^\circ(T_1) + \Delta_\mathrm{r} C_\mathrm{p}^\circ (T_2 - T_1) \tag{5} $$ Eq. (5) is the one you wrote. Typically, $ C_\mathrm{p,j}^\circ$ is given also at the temperature $T = \pu{298.15 K}$ at the back of the books. Again the reasoning is the same, larger the range, worse the estimation. Again, you can combine Eq. (5) and Eq. (1) to obtain an estimate of $K(T_2)$. This calculation often happens, generally for non-mainstream compounds where maybe you only have the heat capacity at $T = \pu{298.15 K}$.
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The equation you wrote assumes that the standard heat of reaction $\Delta H^0$ over the temperature range from T1 to T2 is nearly constant and that A, B, C, and D are ideal gases. If the standard heat of reaction over the temperature range is not nearly constant, an integration over the temperature range must be applied.

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  • $\begingroup$ It does not seem to be assuming an ideal gas. Are you sure you are not confusing it with Clausius–Clapeyron equation of the similar pattern ? That one does. $\endgroup$
    – Poutnik
    Jun 2, 2023 at 12:02
  • $\begingroup$ @Poutnik. Sorry. You are right. $\endgroup$ Jun 2, 2023 at 14:42
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The reason equilibria change with temperature is heat, energy in general, is involved with the reaction. At equilibrium forward and reverse reactions have equal rates and usually different rate constants and activation energies. A Temperature increase raises the energy of the products and reactants, each in different amounts depending on their heat capacities and overall chemical potential energy. This involves chemical study. Generally the lower energy products increase in energy faster so raising the T favors the endothermic reaction direction. Delta H or U does change with T so in detailed study must be considered. For small temperature changes the change in Delta H is usually small.

This is best explained in any good PChem text.

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