# How can we convert DeltaH_f^@ at 298K to DeltaH_f^@ at different temperatures ?? Also tell how to convert bond energy at 298K to bond energy at different temperature

Mar 18, 2017

You'd need to heat the product from standard temperature to nonstandard temperature at the same pressure and add that heating enthalpy.

However, I don't think bond enthalpy changes significantly at a different temperature. We can think of it like an activation energy for that particular molecule's dissociation reaction (in the gas phase to keep it simple):

$\text{AB"(g) -> "A"(g) + "B} \left(g\right)$

Activation energy is almost completely temperature independent, so we can assume that so is bond enthalpy.

Suppose you have the $\Delta {H}_{f}^{\circ}$ for ${\text{NH}}_{3}$ at $\text{298.15 K}$, for the reaction

$3 {\text{H"_2(g) + "N"_2(g) -> 2"NH}}_{3} \left(g\right)$,

and suppose the reactant concentrations were constantly replenished such that the equilibrium is constantly pushed way towards ${\text{NH}}_{3}$.

The $\Delta {H}_{f}^{\circ}$ ($= \Delta {H}_{\text{rxn}}^{\circ}$ since all reactants were in their standard states) for ${\text{NH}}_{3} \left(g\right)$ in this reaction is $- 45.94 \pm 0.35$ $\text{kJ/mol}$ from the NIST webbook.

This is the thermodynamic cycle for this process, to get the big picture: Usually we know $\Delta {H}_{\text{rxn}}^{\circ}$. To fill in $\Delta H$ for heating curves, we have the following formula for heating from ${T}_{1}$ to ${T}_{2}$:

$\Delta {H}_{{T}_{1}}^{{T}_{2}} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{P} \mathrm{dT}$

This requires that you know the constant-pressure heat capacity of
your compound(s) and that you assume that it stays constant in the relevant temperature range.

${C}_{P}$ can be looked up via the NIST webbook. For example, ${\text{NH}}_{3}$ has:

http://webbook.nist.gov/cgi/cbook.cgi?ID=C7664417&Type=JANAFG&Plot=on

a ${C}_{P}$ of $\approx \text{35.64 J/mol"cdot "K}$ at $\text{298.15 K}$ and $\approx \text{42.01 J/mol"cdot"K}$ at $\text{500 K}$.

So, from using the average ${C}_{P}$ in between the two temperatures, heating ${\text{NH}}_{3}$ would approximately contribute a $\Delta H$ of:

$\Delta {H}_{298.15}^{500} = {\int}_{298.15}^{500} \frac{42.01 + 35.64}{2} \mathrm{dT}$

$\approx 38.825 {T}_{2} - 38.825 {T}_{1}$

$= 38.825 \Delta T$

$=$ $38.825 \left(500 - 298.15\right)$

$=$ $\text{7836.83 J/mol}$ $\to$ $\text{7.937 kJ/mol}$

(notice the resemblance to "$q = m {c}_{s} \Delta T$".)

The $\Delta {H}_{f}^{\circ}$ for ${\text{NH}}_{3} \left(g\right)$ was $- 45.94 \pm 0.35$ $\text{kJ/mol}$ from the NIST webbook. So, at $\text{500 K}$, it would be approximately:

$\left(- 45.94 + 7.937\right)$ $\text{kJ/mol}$ $\approx \textcolor{b l u e}{- \text{38.10 kJ/mol}}$

It's less negative at a higher temperature, meaning that it is less exothermic to form ${\text{NH}}_{3} \left(g\right)$ at higher temperatures.

This makes physical sense because we are just assuming the heat released due to making it at a lower temperature is re-absorbed to heat the surroundings further (thus making the enthalpy released less negative).