Thermochemical Equation — Overview & Examples

Before this, knowHeat CapacityHeat of Reaction and Enthalpy

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As we've studied thermodynamics, we've seen a lot of different chemical reactions. Our goal was to calculate the enthalpy or internal energy. We had multiple calculation methods. For example, we've studied Hess's law and heats of formation. But why did we do all those calculations? We wanted to find the reaction enthalpy. It tells us the energy transferred during a chemical reaction. If we know it, we can write a thermochemical equation.

What is a thermochemical equation?

A thermochemical equation is a balanced chemical equation with its heat of reaction. They look like, A+B→C ΔH=XjoulesA + B \rightarrow C\, \Delta H= X joules. You already know how to write and balance chemical equations. So, all you have to do is add the ΔH\Delta H or ΔU\Delta U values.

So, why do we care about thermochemical equations? They give us important information. For example, consider the combustion of gasoline. We can compare the calorimetry value to an engine. That helps determine an engine's efficiency. All car companies are trying to improve fuel efficiency. They compare their cars' engines to the thermochemical data.

Solving Thermochemical Equation Problems

Most thermochemical equation questions are stoichiometry problems. You'll use the mole ratios, molar mass, and density to solve the questions. But, you'll need to consider the energy component.

A new type of problem is efficiency problems. For example, the question might say a process is only sixty percent efficient. So, you'd have to account for the forty percent loss. We'll look at some examples below. As a side note, efficiency problems are an application of the first law of thermodynamics. Energy may be lost from the system to the surroundings. But, the total energy is conserved.

Thermochemical Equation Example Problems

I drive a 2007 ford focus. Its tank holds 14.0 gallons. Let's do some calculations based on my car. The primary fuel in gasoline is octane. It's an organic molecule. Octane has eight carbons and eighteen hydrogens. So, its molecular formula is C8H18C_{8}H_{18}.

Car engines combust octane. The combustions do work by moving the engine's pistons. Through a series of gears, the piston's movement rotates the tires. Unfortunately, internal combustion engines are only about 35% efficient.

Octane's enthalpy of combustion is −10942.0KJ-10942.0 KJ. Its density is 0.703gml0.703 \frac{g}{ml}.

  1. Write the thermochemical equation.
  2. Calculate the enthalpy if I use half a tank of gas.
  3. Find the energy used and lost by the engine from consuming a full tank.
  • To write the thermochemical equation, balance the equation for octane combustion. Add the enthalpy of combustion to the end.
  • 2C8H18(g)+25O2(g)→16CO2(g)+18H2O(g) ΔH=−10942.0KJ2C_{8}H_{18(g)} + 25O_{2(g)} \rightarrow 16CO_{2(g)} + 18H_{2}O_{(g)}\, \Delta H = -10942.0 KJ
  • To calculate the enthalpy for half a tank of gas, we'll need to use the density and the mole ratios. We'll use the density to find the grams of octane burned. Next, we can convert grams to moles. Finally, we can use the mole ratios to solve for the enthalpy. But first, we have to use dimensional analysis to convert gallons to SI units.
  • 7.0 gallons×3785.41mlgallon×0.703gml×1114.23gmol=163.1mols.7.0\,gallons \times 3785.41\frac{ml}{gallon} \times 0.703\frac{g}{ml} \times \frac{1}{114.23\frac{g}{mol}}= 163.1 mols.
  • Now that we found the moles, we can use our thermochemical equation. We see that every two moles of octane release −10942.0KJ-10942.0 KJ.
  • Our molar enthalpy is −10942KJ2mols=−5471KJmol\frac{-10942KJ}{2 mols}= -5471\frac{KJ}{mol}.
  • 163.1mols×−5471KJmol=−892401.7KJ163.1 mols \times -5471\frac{KJ}{mol}= -892401.7 KJ
  • A full tank's energy is double the half tank. So, it releases 1784803.3KJ1784803.3KJ. We said an engine is only 35% efficient. So, the percentage of energy lost is 65%.
  • 1784803.3KJ×0.35=−624681.2KJ used1784803.3KJ \times 0.35 = -624681.2 KJ\,used
  • 1784803.3KJ×0.65=−1160122.1KJ lost1784803.3KJ \times 0.65 = -1 160122.1 KJ\,lost

Video

(Video) Thermochemical Equations Practice Problems

by Tyler DeWitt

In this video, Tyler DeWitt shows you four thermochemical equation example problems. First, you'll learn how to find the total heat released. In the other two examples, you have to do the opposite! You're given the heat and have to find the starting mass. Both types of problems could show up on an exam. So, this video will help you master thermochemical equations.

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