We use cookies to give you a better experience, if that’s ok you can close this message and carry on browsing. For more info read our cookies policy.
2.3

Hanyang University

Skip to 0 minutes and 1 secondThen let's consider the formation reaction at temperatures different from the standard condition, for example at 1000 K. What would be the heat of formation? The heat of reaction is the enthalpy change between after the reaction and before the reaction. Thus the heat of reaction is the difference of enthalpies between products and reactants. Therefore, for the iron oxide formation reaction, the heat of reaction is enthalpy of iron oxide at 1000K - the enthalpy of solid iron at 1000K - half of the enthalpy of oxygen gas at 1000K. In calculating this enthalpy change, we need to know the enthalpy of species i at temperature T. This can be calculated from the enthalpy at standard condition and heat capacity Cp.

Skip to 0 minutes and 58 secondsThe enthalpy at temperature T is summation of enthalpy at 298K and integration of Cp over the temperature of interest. So, enthalpy of iron oxide at 1000K is enthalpy of iron oxide at 298K + integration of Cp of iron oxide from 298K to 1000K. Enthalpy of solid iron and oxygen molecule at 1000K can be written similarly. Rearrange the equation so that the enthalpies at 298K come together. This is the heat of formation of iron oxide at reference state, Hf nought which is tabulated. So, in general, the heat of formation at temperature T can be written like this. It is Hf nought + integration of Cp over temperature for the products - integration of Cp over temperature for the reactants.

Skip to 2 minutes and 7 secondsThe other way to calculate the enthalpy changes at temperature T is to take detour for the corresponding reaction. Instead of directly reacting solid iron and oxygen at 1000K, we can make several steps from the reactants to the final products. First, cooling down the reactants to the room temperature. The enthalpy of cooling for iron and oxygen is delta H1 and delta H2. Then react solid iron and oxygen to generate iron oxide at room temperature. The enthalpy involved is the Hf nought for iron oxide by definition. Finally, heat up iron oxide from room temperature to the final temperature 1000K. The enthalpy of heating is delta H3.

Skip to 2 minutes and 56 secondsThe combination of these three steps is detour for the original reaction, but the initial and final states are the same. Since each reaction is a reversible reaction, it is called the reversible path. Since enthalpy is a state function, its value is path independent, thus the enthalpy of the original reaction is the same with the detour path enthalpy. Thus, Enthalpy of reaction delta H is delta H1 + delta H2 + delta Hf nought for iron oxide + delta H3. Delta H1 and H2 are the simple cooling of solid iron and oxygen. delta H3 is the heating of iron oxide. Thus they can be calculated from the knowledge on the heat capacity, Cp's of the corresponding materials.

Skip to 3 minutes and 56 secondsThe enthalpy change is thus written like this. The result is actually the same with first method for calculation. If the enthalpy change is negative, it is said exothermic and the heat is evolved from the change, here the reaction. If delta h is positive, it is endothermic. Heat must be added to the process. To confirm the concept of reference state and the heat of formation, let's calculate the enthalpy changes in oxidation reaction of diamond at standard condition. Diamond and oxygen react producing CO2 gas. Is the enthalpy change of this reaction the same with heat of formation for CO2? If the reacting elements are in their reference state, then the heat of reaction is the heat of formation for the product.

Skip to 4 minutes and 53 secondsFor carbon material, the reference state is the graphite. So the enthalpy at the standard state is zero for the graphite. For diamond, the enthalpy at the standard state is 1.9 kJ/mole, the non-zero value since it is not the reference state. So, we know that the enthalpy change of this reaction is different from the heat of formation for CO2. To calculate delta H, let's take the reversible path. Let's change the diamond into graphite. The enthalpy change is the difference of enthalpy between graphite and diamond, thus -1.9 kJ/mole. Then react graphite with oxygen. The heat of oxidation reaction of graphite is the heat of formation for CO2 The overall enthalpy changes is the sum of those two.

Heat of formation at 1000 K

We can calculate the enthalpy changes at temperatures other than reference temperature by taking a reversible detour.

Integration of ${ C }_{ p }$ over temperature gives the energy changes upon temperature change within a single phase. This energy change under constant pressure is called sensible heat. When there are phase transitions during the temperature range of interest, the latent heat should be considered also.