# 5.3 Enthalpy  (Page 7/25)

 Page 7 / 25

## Writing reaction equations for $\text{Δ}{H}_{\text{f}}^{°}$

Write the heat of formation reaction equations for:

(a) C 2 H 5 OH( l )

(b) Ca 3 (PO 4 ) 2 ( s )

## Solution

Remembering that $\text{Δ}{H}_{\text{f}}^{°}$ reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:

(a) $2\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)+3{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}\left(l\right)$

(b) $3\text{Ca}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{P}}_{4}\left(s\right)+4{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Ca}}_{3}\left({\text{PO}}_{4}{\right)}_{2}\left(s\right)$

Note: The standard state of carbon is graphite, and phosphorus exists as P 4 .

Write the heat of formation reaction equations for:

(a) C 2 H 5 OC 2 H 5 ( l )

(b) Na 2 CO 3 ( s )

(a) $4\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)+5{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}{\text{OC}}_{2}{\text{H}}_{5}\left(l\right);$ (b) $2\text{Na}\left(s\right)+\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Na}}_{2}{\text{CO}}_{3}\left(s\right)$

## Hess’s law

There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.

This type of calculation usually involves the use of Hess’s law    , which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps . Hess’s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:

$\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=-394\phantom{\rule{0.2em}{0ex}}\text{kJ}$

In the two-step process, first carbon monoxide is formed:

$\text{C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=-111\phantom{\rule{0.2em}{0ex}}\text{kJ}$

Then, carbon monoxide reacts further to form carbon dioxide:

$\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=-283\phantom{\rule{0.2em}{0ex}}\text{kJ}$

The equation describing the overall reaction is the sum of these two chemical changes:

$\begin{array}{}\\ \text{Step 1: C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)\\ \underset{¯}{\text{Step 2: CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)}\\ \text{Sum: C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)+\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)+{\text{CO}}_{2}\left(g\right)\end{array}$

Because the CO produced in Step 1 is consumed in Step 2, the net change is:

$\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)$

According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps. We can apply the data from the experimental enthalpies of combustion in [link] to find the enthalpy change of the entire reaction from its two steps:

$\begin{array}{ll}\text{C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)\hfill & \text{Δ}{H}_{298}^{°}=-111\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \\ \frac{\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)}{\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{1em}{0ex}}}\hfill & \frac{\text{Δ}{H}_{298}^{°}=-283\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{Δ}{H}_{298}^{°}=-394\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$

The result is shown in [link] . We see that Δ H of the overall reaction is the same whether it occurs in one step or two. This finding (overall Δ H for the reaction = sum of Δ H values for reaction “steps” in the overall reaction) is true in general for chemical and physical processes.

what is an atom
An atom is the smallest particle of an element which can take part in a chemical reaction..
olotu
hello
Karan
how to find the rate of reaction?
Karan
what is isomerism ?
Formula for equilibrium
is it equilibrium constant
olotu
Yes
Olatunde
yes
David
what us atomic of molecule
chemical formula for water
H20
Samson
what is elemental
what are the properties of pressure
Maryam
How can water be turned to gas
VICTOR
what's a periodic table
how does carbon catenate?
condition in cracking from Diesel to petrol
hey I don't understand anything in chemistry so I was wondering if you could help me
i also
Okikiola
I also
Brient
hello
Brient
condition for cracking diesel to form kerosene
Brient
Really?
Isa
yes
Brient
can you tell me
Brient
Brient
what is periodic law
periodic law state that the physical and chemical properties of an element is the periodic function of their atomic number
rotimi
how is valency calculated
How is velency calculated
Bankz
Hi am Isaac, The number of electrons within the outer shell of the element determine its valency . To calculate the valency of an element(or molecule, for that matter), there are multiple methods. ... The valency of an atom is equal to the number of electrons in the outer shell if that number is fou
YAKUBU
what is the oxidation number of this compound fecl2,fecl3,fe2o3
bonds formed in an endothermic reaction are weaker than the reactants but y r these compound stable at higher temperatures
what is a disproportionation reaction