Question Video: Calculating the Standard Enthalpy of Formation for Heptane Using Standard Enthalpies of Combustion | Nagwa Question Video: Calculating the Standard Enthalpy of Formation for Heptane Using Standard Enthalpies of Combustion | Nagwa

# Question Video: Calculating the Standard Enthalpy of Formation for Heptane Using Standard Enthalpies of Combustion Chemistry • First Year of Secondary School

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Using Hess’s law and the standard enthalpies of combustion in the given table, what is the standard enthalpy of formation of heptane (C₇H₁₆(l))?

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### 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.

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