Video Transcript
Using Hess’s law and the standard
enthalpies of combustion in the given table, what is the standard enthalpy of
formation of heptane, C7H16, liquid?
It can be difficult to measure
enthalpy changes for dangerous or slow reactions. For situations like this, we can
calculate enthalpy changes indirectly by using enthalpy data from other
reactions. In this question, we have been
asked to determine the standard enthalpy of formation for heptane when given the
standard enthalpies of combustion for several substances.
The standard enthalpy of formation
is the enthalpy change when one mole of substance forms from its constituent
elements in their standard states under standard conditions. The standard enthalpy of combustion
is the enthalpy change when one mole of substance burns completely in oxygen under
standard conditions and standard states. But how can we use the standard
enthalpies of combustion given in the table to calculate the standard enthalpy of
formation?
We can use a Hess cycle to help
us. Let’s clear some space to work. Let’s start by writing an equation
for the formation of heptane from its constituent elements in their standard
states. We can write C solid plus H2 gas
react to form C7H16 liquid. To balance this equation, we should
add a coefficient of seven in front of carbon and a coefficient of eight in front of
diatomic hydrogen. Let’s label this reaction one. The enthalpy change of this
reaction is the standard enthalpy of formation of heptane.
Now, we need to write equations for
the combustion of carbon, diatomic hydrogen, and heptane. The combustion of carbon in excess
oxygen will produce carbon dioxide, and the combustion of hydrogen will produce
water. Again, we must ensure our equation
is balanced. Let’s use a coefficient of seven in
front of CO2 and a coefficient of eight in front of water. Let’s label this reaction two and
note that the direction of the arrow is towards the products of combustion.
Next, let’s consider the combustion
of heptane. When combusted in excess oxygen,
heptane will also produce seven moles of carbon dioxide and eight moles of
water. We can label this reaction three
and again note that the direction of the arrow is towards the products of
combustion.
Now we have an alternative way to
calculate the standard enthalpy of formation of heptane using the standard
enthalpies of combustion given in the table. We can state that the enthalpy
change of reaction one is equal to the enthalpy change of reaction two plus the
enthalpy change of reaction three. However, it’s important to note
that when taking this alternative route, we must go in the opposite direction of the
arrow in reaction three. Therefore, we must change the sign
of the enthalpy value of reaction three when performing the calculation.
To calculate the enthalpy change
for reaction two, we need to take seven times the standard enthalpy of combustion of
solid carbon plus eight times the standard enthalpy of combustion of H2 gas. Seven and eight are the
coefficients from the balanced chemical equation. After substituting in the values
from the table for carbon and diatomic hydrogen, we find that the enthalpy change of
reaction two is negative 5046 kilojoules per mole.
Now, let’s consider reaction
three. The enthalpy change of reaction
three is simply the standard enthalpy of combustion of liquid heptane, which
according to the table is negative 4817 kilojoules per mole.
Now we can substitute the values
for the enthalpy changes of reactions two and three into our equation. After substituting and changing the
sign of the enthalpy change of reaction three, we get negative 5046 kilojoules per
mole plus 4817 kilojoules per mole. This gives us an answer of negative
229 kilojoules per mole.
In conclusion, the standard
enthalpy of formation of heptane is negative 229 kilojoules per mole.
