Enthalpy is a state function, but its value depends on temperature. Consequently, reaction enthalpies, formation enthalpies, and other thermodynamic quantities are generally different at different temperatures.
The key to understanding this temperature dependence is the constant-pressure heat capacity:
\[ C_P = \left( \frac{\partial H} {\partial T} \right)_P \]
This definition states that the heat capacity measures how rapidly the enthalpy changes with temperature at constant pressure.
Rearranging gives the differential form
\[ dH = C_P\,dT \]
which allows changes in enthalpy to be calculated from temperature changes.
If the heat capacity remains approximately constant over the temperature range of interest, the expression can be integrated directly:
\[ \Delta H = \int_{T_1}^{T_2} C_P\,dT \]
giving
\[ \Delta H = C_P \int_{T_1}^{T_2} dT \]
\[ \Delta H = C_P (T_2-T_1) \]
\[ \Delta H = C_P\Delta T \]
This approximation is often adequate for modest temperature ranges and is widely used in introductory thermodynamic calculations.
Over larger temperature ranges, heat capacities often vary with temperature. A common empirical model is
\[ C_P = a + bT + \frac{c}{T^2} \]
where \(a\), \(b\), and \(c\) are experimentally determined constants.
The enthalpy change is then calculated by integrating:
\[ \Delta H = \int_{T_1}^{T_2} \left( a+bT+\frac{c}{T^2} \right)dT \]
Carrying out the integration gives
\[ \Delta H = a(T_2-T_1) + \frac{b}{2} \left( T_2^2-T_1^2 \right) - c \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
This expression allows enthalpy changes to be calculated over broad temperature ranges using experimentally measured heat-capacity data.
Reaction enthalpies also depend on temperature because the reactants and products have different heat capacities.
The temperature dependence of a reaction enthalpy is given by
\[ \Delta H_{\mathrm{rxn}}(T_2) = \Delta H_{\mathrm{rxn}}(T_1) + \int_{T_1}^{T_2} \Delta C_P\,dT \]
where
\[ \Delta C_P = \sum_{\mathrm{products}} \nu_iC_{P,i} - \sum_{\mathrm{reactants}} \nu_iC_{P,i} \]
If \(\Delta C_P\) is approximately constant,
\[ \Delta H_{\mathrm{rxn}}(T_2) = \Delta H_{\mathrm{rxn}}(T_1) + \Delta C_P (T_2-T_1) \]
This relationship allows thermodynamic data measured at one temperature to be adjusted to another temperature.
Big picture: Heat capacities connect temperature changes to enthalpy changes. By integrating heat-capacity expressions, we can determine how the enthalpies of substances and reactions vary with temperature.
A \(25.0\ \mathrm{g}\) sample of liquid water is heated from \(18.0^\circ\mathrm{C}\) to \(45.0^\circ\mathrm{C}\). Calculate the enthalpy change for the process.
For liquid water, the specific heat capacity is
\[ C_P = 4.184\ \mathrm{J\,g^{-1}\,K^{-1}} \]
Since the heat capacity is approximately constant over this temperature range,
\[ \Delta H = mC_P\Delta T \]
First calculate the temperature change:
\[ \Delta T = 45.0^\circ\mathrm{C} - 18.0^\circ\mathrm{C} \]
\[ \Delta T = 27.0\ \mathrm{K} \]
Substituting the values:
\[ \Delta H = (25.0\ \mathrm{g}) \left( 4.184\ \mathrm{J\,g^{-1}\,K^{-1}} \right) (27.0\ \mathrm{K}) \]
\[ \Delta H = 2824\ \mathrm{J} \]
\[ \Delta H = 2.82\ \mathrm{kJ} \]
Therefore,
\[ \boxed{ \Delta H = +2.82\ \mathrm{kJ} } \]
Physical interpretation: Because the water is heated, energy must flow into the sample. The positive value of \(\Delta H\) indicates that the process is endothermic and requires \(2.82\ \mathrm{kJ}\) of energy to raise the temperature of the water from \(18.0^\circ\mathrm{C}\) to \(45.0^\circ\mathrm{C}\).
A substance has a temperature-dependent heat capacity given by
\[ C_P = 2.00\ \mathrm{J\,g^{-1}\,K^{-1}} - 1.00\times10^{-3} \ \mathrm{J\,g^{-1}\,K^{-2}} (T) \]
Calculate the enthalpy change when a \(50.0\ \mathrm{g}\) sample is heated from \(283\ \mathrm{K}\) to \(363\ \mathrm{K}\).
Since the heat capacity depends on temperature, we must integrate:
\[ \Delta H = m \int_{T_1}^{T_2} C_P(T)\,dT \]
Substituting the expression for \(C_P\):
\[ \Delta H = (50.0\ \mathrm{g}) \int_{283}^{363} \left( 2.00 - 1.00\times10^{-3}T \right) dT \]
Evaluating the integral:
\[ \Delta H = (50.0\ g) \left[ 2.00T - \frac{1.00\times10^{-3} \frac{J}{g\ K^2}}{2} T^2 \right]_{283\ K}^{363\ K} \]
\[ \Delta H = (50.0\ g) \left[ 2.00 \frac{J}{g\ K}(363\ K) - \frac{1.00\times10^{-3} \frac{J}{g\ K^2}}{2} (363\ K)^2 \right. \]
\[ \left. - \left( 2.00 \frac{J}{g\ K}(283\ K) - \frac{1.00\times10^{-3} \frac{J}{g\ K^2}}{2} (283\ K)^2 \right) \right] \]
\[ \Delta H = (50.0\ g) (54.16\ \frac{J}{g}) \]
\[ \Delta H = 2.71\times10^3\ \mathrm{J} \]
\[ \Delta H = 2.71\ \mathrm{kJ} \]
Therefore,
\[ \boxed{ \Delta H = +2.71\ \mathrm{kJ} } \]
Physical interpretation: The heat capacity decreases slightly as the temperature increases, so less energy is required to raise the temperature near \(363\ \mathrm{K}\) than near \(283\ \mathrm{K}\). As a result, the enthalpy change is slightly smaller than would be predicted using a constant heat capacity of \(2.00\ \mathrm{J\,g^{-1}\,K^{-1}}\).
Consider the reaction
\[ 2Na(s) + Cl_2(g) \rightarrow 2NaCl(s) \]
The standard reaction enthalpy at \(25^\circ\mathrm{C}\) (\(298\ \mathrm{K}\)) is
\[ \Delta H^\circ_{\mathrm{rxn}} (298\ \mathrm{K}) = -822.4\ \mathrm{kJ} \]
Calculate the reaction enthalpy at \(95^\circ\mathrm{C}\) (\(368\ \mathrm{K}\)).
The heat-capacity change for the reaction is
\[ \Delta C_P = \sum \nu C_{P,\mathrm{products}} - \sum \nu C_{P,\mathrm{reactants}} \]
\[ \Delta C_P = 2(50.46 \frac{J}{mol\ K}) - \left[ 2(28.23 \frac{J}{mol\ K}) + 33.95 \frac{J}{mol\ K} \right] \]
\[ \Delta C_P = 10.51\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
Assuming \(\Delta C_P\) is constant,
\[ \Delta H_{\mathrm{rxn}}(T_2) = \Delta H_{\mathrm{rxn}}(T_1) + \Delta C_P (T_2-T_1) \]
\[ \Delta H_{\mathrm{rxn}}(368) = -822.4\ \mathrm{kJ} + \left( 10.51\ \mathrm{J\,mol^{-1}\,K^{-1}} \right) (368\ K - 298\ K) \]
Converting the correction to kJ:
\[ \Delta H_{\mathrm{corr}} = \frac{ (10.51 \frac{J}{mol\ K})(70\ K) } {1000 \frac{kJ}{J}} = 0.736\ \mathrm{kJ} \]
\[ \Delta H_{\mathrm{rxn}}(368) = -822.4 \frac{kJ}{mol} + 0.736 \frac{kJ}{mol} \]
\[ \Delta H_{\mathrm{rxn}}(368) = -821.7\ \mathrm{kJ} \]
| Temperature | \(\Delta H^\circ_{\mathrm{rxn}}\) |
|---|---|
| \(25^\circ\mathrm{C}\) | \(-822.4\ \mathrm{kJ}\) |
| \(95^\circ\mathrm{C}\) | \(-821.7\ \mathrm{kJ}\) |
Physical interpretation: The products have a slightly larger total heat capacity than the reactants, causing the reaction enthalpy to become slightly less negative as the temperature increases. Although the change is small over a \(70\ \mathrm{K}\) interval, the effect becomes increasingly important over larger temperature ranges.
Choose the correct answer for the randomly generated problem.
Big picture: Heat capacities connect temperature changes to enthalpy changes. By integrating heat-capacity expressions, we can determine how the enthalpies of substances and chemical reactions vary with temperature, allowing thermodynamic data measured at one temperature to be applied at another.