Video Transcript
The equation shows the formation of
one mole of ethanol, C2H5OH, from its constituent elements under standard conditions
and with standard states. Two C solid plus three H2 gas plus
three and one-half O2 gas react to form CH3CH2OH liquid plus three O2 gas. Given the combustion values in the
table, calculate the value for the standard enthalpy of formation.
It can be difficult to measure
enthalpy changes for dangerous or slow reactions. For reactions like these, we can
calculate enthalpy changes indirectly by using enthalpy data from other
reactions. In this question, we have been
given a chemical equation showing the formation of ethanol from its constituent
elements. Let’s clear some space and rewrite
this balanced equation below.
Let’s label this reaction as
reaction one. The enthalpy change for this
reaction corresponds to the standard enthalpy of formation of ethanol. By definition, the standard
enthalpy of formation is the enthalpy change when one mole of substance forms from
its elements in their standard states under standard conditions. Unfortunately, measuring the
enthalpy change of this reaction is difficult. In order to calculate the standard
enthalpy of formation of ethanol, we will need to construct a Hess cycle and use the
standard enthalpies of combustion provided in the table. Let’s clear some space to define
the standard enthalpy of combustion.
The standard enthalpy of combustion
is the enthalpy change when one mole of substance burns completely in oxygen under
standard conditions and standard states. The combustion of pure carbon in
excess oxygen will produce carbon dioxide. The combustion of hydrogen will
result in the production of water. Since combustion refers to the
burning of a substance in oxygen, we can ignore the standard enthalpy of combustion
of oxygen.
Let’s label this reaction as
reaction two and ensure that the equation is balanced. We need to write a coefficient of
two in front of carbon dioxide and a coefficient of three in front of water. The enthalpy change of reaction two
corresponds to the standard enthalpy of combustion of two moles of carbon plus the
standard enthalpy of combustion of three moles of diatomic hydrogen. We need to multiply the standard
enthalpies of combustion by two and three, which are the coefficients from the
balanced chemical equation.
Next, let’s consider the combustion
of ethanol. When ethanol combusts, it will also
produce two moles of carbon dioxide and three moles of water. Let’s label this reaction as
reaction three. The enthalpy change of reaction
three corresponds to the standard enthalpy of combustion of ethanol. Now, we should follow the Hess
cycle from the reactants to the products of reaction one. So, we will complete reaction two
in the direction that it is written but complete reaction three in the opposite
direction of the reaction arrow. We can therefore state that the
enthalpy change of reaction one is equal to the enthalpy change of reaction two
minus the enthalpy change of reaction three.
Now we can use the expressions we
wrote to help us substitute the values from the table into the pink equation. For Δ𝐻 two, we should add two
times the standard enthalpy of combustion of carbon to three times the standard
enthalpy of combustion of hydrogen. Δ𝐻 three is equal to the standard
enthalpy of combustion of ethanol, which is negative 1,371 kilojoules per mole. After simplification, we find that
Δ𝐻 one equals negative 1,646 kilojoules per mole plus 1,371 kilojoules per
mole. After adding, we get an answer of
negative 275 kilojoules per mole.
In conclusion, the standard
enthalpy of formation of ethanol is negative 275 kilojoules per mole.
