# Table of Some Standard Enthalpies of Formation at 25°C

Submitted by ChemPRIME Staff on Thu, 12/09/2010 - 14:02

### Text below taken from Hess' Law

Perhaps the most useful feature of thermochemical equations is that they can be combined to determine ΔHm values for other chemical reactions. Consider, for example, the following two-step sequence. Step 1 is reaction of 1 mol C(s) and 0.5 mol O2(g) to form 1 mol CO(g):

C(s) + ½O2(g) → CO(g)      ΔHm = –110.5 kJ = ΔH1

(Note that since the equation refers to moles, not molecules, fractional coefficients are permissible.) In step 2 the mole of CO reacts with an additional 0.5 mol O2 yielding 1 mol CO2:

CO(g) + ½O2(g) → CO2(g)      ΔHm = –283.0 kJ = ΔH2

The net result of this two-step process is production of 1 mol CO2 from the original 1 mol C and 1 mol O2 (0.5 mol in each step). All the CO produced in step 1 is used up in step 2.

On paper this net result can be obtained by adding the two chemical equations as though they were algebraic equations. The CO produced is canceled by the CO consumed since it is both a reactantA substance consumed by a chemical reaction. and a product of the overall reaction

Experimentally it is found that the enthalpyA thermodynamic state function, symbol H, that equals internal energy plus pressure x volume; the change in enthalpy corresponds to the energy transferred as a result of a temperature difference (heat transfer) when a reaction occurs at constant pressure. change for the net reaction is the sum of the enthalpy changes for steps 1 and 2:

ΔHnet = –110.5 kJ + (–283.0 kJ) = –393.5 kJ = ΔH1 + ΔH2

That is, the thermochemical equation

C(s) + O2(g) → CO2(g)      ΔHm = –393.5 kJ

is the correct one for the overall reaction.

In the general case it is always true that whenever two or more chemical equations can be added algebraically to give a net reaction, their enthalpy changes may also be added to give the enthalpy change of the net reaction.

This principle is known as Hess' lawThe generalization that if a chemical reaction can be written as the sum of two or more other chemical reactions, then the heat of the overall reaction is the sum of the heats of reaction for the other reactions. A similar rule applies to enthalpy and Gibbs energy.. If it were not true, it would be possible to think up a series of reactions in which energyA system's capacity to do work. would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain ΔHm values for reactions which cannot be carried out experimentally, as the next example shows.

EXAMPLE 1 Acetylene (C2H2) cannot be prepared directly from its elements according to the equation

2C(s) + H2(g) → C2H2(g)      (1)

Calculate ΔHm for this reaction from the following thermochemical equations, all of which can be determined experimentally:

C(s) + O2(g) → CO2(g)      ΔHm = –393.5 kJ  (2a)

H2(g) + ½O2(g) → H2O(l)      ΔHm = –285.8 kJ   (2b)

C2H2(g) + $\tfrac{\text{5}}{\text{2}}$O2(g) → 2CO2(g) + H2O(l)      ΔHm = –1299.8 kJ    (2c)

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (1):

a) Since Eq. (1) has 2 mol C on the left, we multiply Eq. (2a) by 2.

b) Since Eq. (1) has 1 mol H2 on the left, we leave Eq. (2b) unchanged.

c) Since Eq. (1) has 1 mol C2H2 on the right, whereas there is 1 mol C2H2 on the left of Eq. (2c) we write Eq. (2c) in reverse.

We then have

$\text{2 C(s) + O}_{\text{2}}\text{(g)}~\to~ \text{CO}_{\text{2}}\text{(g)}~~~~~~~~~~~~~~~~~~~~~\Delta \text{H}_{\text{m}}=\text{ 2 (-393.5) kJ}$

$\text{H}_{\text{2}}\text{(g) + }\tfrac{1}{2}\text{O}_{\text{2}}\text{(g)}~\to~ \text{H}_{\text{2}}\text{O(l)}~~~~~~~~~~~~~~~~~~~~~\Delta \text{H}_{\text{m}} = \text{-285.8 kJ}$

$\underline{\text{2 CO}_{\text{2}}\text{(g) + H}_{\text{2}}\text{O}(l)}\underline{\to \text{ C}_{\text{2}}\text{H}_{\text{2}}\text{(g) +}\tfrac{5}{2}\text{O}_{\text{2}}\text{(g)}}~~~~\Delta H_{\text{m}}=-\text{(}-\text{1299.8 kJ)}$

$\text{2 C(s) + H}_{\text{2}}\text{(g) + 2}\tfrac{1}{2}\text{O}_{\text{2}}\text{(g)}~\to~ \text{C}_{\text{2}}\text{H}_{\text{2}}\text{(g) +}\tfrac{5}{2}\text{O}_{\text{2}}\text{(g)}$

ΔHm = (-787.0 -285.8 + 1299.8) kJ

=227.0 kJ

Cancelling 5/2 O2 on each side, the desired result is

2C(s) + H2(g) → C2H2(g)      ΔHm = 227.0 kJ