|Thermal equations||Hess's law energy level diagram|
A + 2 B = AB2, dH1 2 = dH1 + dH2
======= A + 2 B | | dH1 dH1 2 | ===== AB + B | | dH2 ======= AB2
If both sides of the equations are multiplied by a factor to alter the number of moles, dH, dH°, or dHf for the equation should be multiplied by the same factor, since they are quantities per equation as written. Thus, for the equation
For the reverse reaction, the sign of these quantities are changed (multiply by -1). The equation implies the following:
Hess's law states that energy changes are state functions. The amount of energy depends only on the states of the reactants and the state of the products, but not on the intermediate steps. Energy (enthalpy) changes in chemical reactions are the same, regardless whether the reactions occur in one or several steps. The total energy change in a chemical reaction is the sum of the energy changes in its many steps leading to the overall reaction.
For example, in the diagram below, we look at the oxidation of carbon into CO and CO2. The direct oxidation of carbon (graphite) into CO2 yields an enthalpy of -393 kJ/mol. When carbon is oxidized into CO and then CO is oxidized to CO2, the enthalpies are -110 and -283 kJ/mol respectively. The sum of enthalpy in the two steps is exactly -393 kJ/mol, same as the one-step reaction.
0 kJ ------------ C(graphite) + O2 | | -110 kJ | | V | CO + 0.5 O2 ----- | | | -393 kJ | | -283 kJ | | | | V V ------------ CO2The two-step reactions are:
Hess's law can be applied to calculate enthalpies of reactions that are difficult to measure. In the above example, it is very difficult to control the oxidation of graphite to give pure CO. However, enthalpy for the oxidation of graphite to CO2 can easily be measured. So can the enthalpy of oxidation of CO to CO2. The application of Hess's law enables us to estimate the enthalpy of formation of CO. Since,
Application of Hess's law enables us to calculate dH, dH°, and dHf for chemical reactions that impossible to measure, providing that we have all the data of related reactions.
Some more examples are given below to illustrate the applications of Hess Law.
Lets interpret the information about enthalpy of formation by writing out the equations:
|(1) H2(g) + 0.5 O2(g) -> H2O(l)||-285.8|
|(2) C(graphite) + O2(g) -> CO2(g)||-293.5|
|(3) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)||-890.4|
From the above equations, derive|
C + 2H2 -> CH4
|Answer: C + 2H2 -> CH4||-74.7|
Hint: 2*(1) + (2) - (3), Thus,|
dHf = 2 * (-285.8) + (-393.5) - (-890.4) = ?
Three enthalpies of reactions involved in this Example are standard enthalpies of formation, and one is the enthalpy of combustion. The formation of methane from graphite and hydrogen cannot be achieved easily, and its enthalpy of formation is not directly measurable, but the calculations like this provide the data to be included in thermodynamic data. The value of -74.4 kJ/mol has been listed in several data sources.
From these data, we can construct an energy level diagram for these chemical combinations as follows:
===C(graphite) + 2 H2(g) + 2 O2(g)=== - 74.7 kJ | | == CH4 (g) + 2 O2(g)== | | | | | | | | |-965.1 kJ -890.4 kJ | | [(-2*285.8-393.5) kJ] | | | | | | | | V V ==========CO2(g) + 2 H2O(l)==========
H2O(l) -> H2O(g) dHo = 44 kJ/mol at 298 K
Add the two equations to give the third one:
Your turn to work:
Compare to result of Example 2, this result differ slightly, due to
a different set of data being used. The difference is 0.2 %.
One method is to re-write the key equations as follows and then add them to
cancel out undesirable compound on both sides. Practice the cancellation
of the formula yourself.
A higher amount of energy (890 vs 802 kJ/mol) is extracted if the exhaust is condensed to liquid water. The exhaust from high efficiency furnace is at lower temperature, and the water vapour is condensed to liquid. However, there is always some lost in a furnace operation.
In order to show how the chemical reactions take place, and for a better appreciation of the technique of problem solving, we write the equations according to the data given:
S(s) + 3/2 O2 -> SO3 dH = -396 kJ
Sketch an energy level diagram for the combinations of substances.
-2(1) + 2(3) + (4) - (6) - 2(7) gives
dH = -2*(-249) + 2*(-803) + (-643) - (-75) - 2(-436)
= -804 kJ/mol
Work out the details yourself and check the result. The calculation is rather complicated. Reading it will not be able to master the technique. Data from equations 2, 5 and 8 are not required. Often, you have to select suitable data from a Table of Standard Enthalpy of Formation in problem solving.
C(graphite) + 2 O(g) -> CO2(g) dHo = -643 kJ/mol
2 O2(g) -> 4 O(g) dHo = 498 kJ/mol
4 H(g) + 2 O(g) -> 2 H2O(g) dHo = -1606 kJ/mol
2 H2(g) -> 4 H(g) dHo = 872 kJ/mol
add all equations ---------------------------add all dHs
CH4 + 2 O2(g) -> CO2(g) + 2 H2O(g) dHo = -804 kJ/mol
Confidence Building Questions
More Hess's law problems
Problem Generator for the application of Hess's law.
Your turn to work:
Compare to result of Example 2, this result differ slightly, due to a different set of data being used. The difference is 0.2 %.
One method is to re-write the key equations as follows and then add them to cancel out undesirable compound on both sides. Practice the cancellation of the formula yourself.