Thermal equations | Hess's law energy level diagram |
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then, A + 2 B = AB_{2}, dH_{1 2} = dH_{1} + dH_{2} |
======= A + 2 B | | dH_{1} dH_{1 2} | ===== AB + B | | dH_{2} ======= AB_{2} |
If both sides of the equations are multiplied by a factor to alter the number of moles, dH, dH°, or dH_{f} 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 CO_{2}. The direct oxidation of carbon (graphite) into CO_{2} yields an enthalpy of -393 kJ/mol. When carbon is oxidized into CO and then CO is oxidized to CO_{2}, 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) + O_{2} | | -110 kJ | | V | CO + 0.5 O_{2} ----- | | | -393 kJ | | -283 kJ | | | | V V ------------ CO_{2}The 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 CO_{2} can easily be measured. So can the enthalpy of oxidation of CO to CO_{2}. 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 dH_{f} 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.
Example 1
Solution:
Lets interpret the information about enthalpy of formation
by writing out the equations:
dH°_{f} /(kJ/mol) | |
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(1) H_{2}(g) + 0.5 O_{2}(g) -> H_{2}O(l) | -285.8 |
(2) C(graphite) + O_{2}(g) -> CO_{2}(g) | -293.5 |
(3) CH_{4}(g) + 2O_{2}(g) -> CO_{2}(g) + 2H_{2}O(l) | -890.4 |
From the above equations, derive C + 2H_{2} -> CH_{4} | |
Answer: C + 2H_{2} -> CH_{4} | -74.7 |
Hint: 2*(1) + (2) - (3), Thus, dH_{f} = 2 * (-285.8) + (-393.5) - (-890.4) = ? |
Discussion:
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 H_{2}(g) + 2 O_{2}(g)=== - 74.7 kJ | | == CH_{4 }(g) + 2 O_{2}(g)== | | | | | | | | |-965.1 kJ -890.4 kJ | | [(-2*285.8-393.5) kJ] | | | | | | | | V V ==========CO_{2}(g) + 2 H_{2}O(l)==========
Example 2
Solution:
Add the two equations to give the third one:
Discussion:
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.
Example 3
Solution:
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:
Your turn to work:
Sketch an energy level diagram for the combinations of substances.
Example 4
Solution:
-2(1) + 2(3) + (4) - (6) - 2(7) gives
Discussion:
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.
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.