Calorific Value(Heat of combustion)

Main Idea

The Calorific Value of a sample, also known as its Heat of Combustion, is defined as the amount of heat released during the complete combustion of the sample. In this chemical reaction, a molecule usually consisting of carbon, hydrogen, nitrogen or oxygen, is usually oxidized by diatomic oxygen to form carbon dioxide and other products. This process is exothermic, and thus releases heat when more stable bonds are formed. The bonds in the reactant are unreactive with oxygen under normal conditions, so the necessary activation energy must be applied. Once the activation energy is provided, the reaction occurs very quickly and violently due to the relative stabilities of the products' bonds. Since combustion is exothermic, one reaction on the molecular level provides the energy necessary to activate another, meaning combustions are often chain reactions and can sustain themselves as long as oxygen is present.

The Caloric Value itself measures the thermal energy released per unit of mass or moles depending on the size of the reaction. Hence, its unit is J/kg, but can be measured in calories per mass as well, as seen in food. This is the SI unit for unit and is defined as being the energy necessary to heat a gram of water by 1 degree Celsius. Since this is a small amount of energy, the calorie is measured in kCal, or Calories with a capital c, just as energy is often represented in kJ. The heat of combustion can be measured either by the higher heat value (HHV) which analyzes simply the energy released by the reaction, and the lower heat value, which incorporates the energy absorbed by water to change from liquid phase to gas phase. If the reaction does not form water, there is no difference between the two values.

The HHV is typically measured through experiments using a bomb calorimeter, where the sample is supplied with excess oxygen. This device measures the temperature change. From this, the Heat of Combustion can be computed using the Thermal Energy Equation.

The LHV is the HHV minus the energy the water absorbed to transition states.

Mathematical Model

A typical combustion reaction looks like this:

[math]\displaystyle{ \text{C}_x\text{H}_y\text{N}_z\text{O}_n + \text{O}_2 \longrightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} + \frac{z}{2}\text{N}_2 \quad Q = \boldsymbol{\Omega} \ \frac{J}{mol} }[/math], where

[math]\displaystyle{ \bullet \ \text{C}_x = }[/math] [math]\displaystyle{ x }[/math] atoms of Carbon

[math]\displaystyle{ \bullet \ \text{H}_y = }[/math] [math]\displaystyle{ y }[/math] atoms of Hydrogen

[math]\displaystyle{ \bullet \ \text{N}_z = }[/math] [math]\displaystyle{ z }[/math] atoms of Nitrogen

[math]\displaystyle{ \bullet \ \text{O}_n = }[/math] [math]\displaystyle{ n }[/math] atoms of Oxygen gas

[math]\displaystyle{ \bullet \ x\text{CO}_2 = }[/math] [math]\displaystyle{ x }[/math] moles of Carbon Dioxide

[math]\displaystyle{ \bullet \ \frac{y}{2}\text{H}_2\text{O} = }[/math] [math]\displaystyle{ \frac{y}{2} }[/math] moles of Water

[math]\displaystyle{ \bullet \ \frac{z}{2}\text{N}_2 = }[/math] [math]\displaystyle{ \frac{z}{2} }[/math] moles of Nitrogen gas

[math]\displaystyle{ \bullet \ Q = }[/math] the Heat of Combustion

[math]\displaystyle{ \bullet \ \boldsymbol{\Omega} = }[/math] a constant

Let us look at this combustion reaction as an example:

[math]\displaystyle{ \text{CH}_{3}\text{OH} + \text{O}_2 \longrightarrow \text{CO}_2 + 2\text{H}_{2}\text{O} \quad Q = 890 \ \frac{kJ}{mol} }[/math]

Here we see the Heat of Combustion of [math]\displaystyle{ \text{CH}_{3}\text{OH} }[/math], Methanol, is [math]\displaystyle{ 890 \ \frac{kJ}{mol} }[/math].

It is also always important to keep the Thermal Energy Equation in mind when thinking of these things, since it does relate Heat to a Temperature change:

[math]\displaystyle{ \Delta Q = mc \Delta T }[/math]

Hess's law can be used to determine the change in enthalpy of a given chemical reaction. By finding the difference between the sums of the bond enthalpies of the products by the sum of the bond enthalpies of the reactants, one can determine how much energy is released or absorbed based on the reaction.

Computational Model

Insert Computational Model Here

Examples

Simple

A 1.55 gram sample of Ethanol is burned in a bomb calorimeter:

[math]\displaystyle{ \text{C}_{2}\text{H}_{5}\text{OH} + 3\text{O}_2 \longrightarrow 2\text{CO}_2 + 3\text{H}_{2}\text{O} }[/math]

a) If this combustion caused a Temperature increase of 55°C in a 200 gram sample of water, what is the Molar Heat of Combustion of the Ethanol? Water has a Specific Heat Capacity of [math]\displaystyle{ C = 4.18 \ \frac{J}{g \cdot °C} }[/math]. Ethanol has a molar mass of [math]\displaystyle{ Mm = 46.1 \ \frac{g}{mol} }[/math].

Using dimensional analysis, we can find the number of moles of Ethanol in the 1.55 gram sample. Letting [math]\displaystyle{ m }[/math] be the mass of the sample in grams, [math]\displaystyle{ Mm }[/math] be the molar mass of Ethanol in grams per mole, and [math]\displaystyle{ M }[/math] be the number of moles of Ethanol, we see:

[math]\displaystyle{ M = \frac{m}{1} \times \frac{1}{Mm} = \frac{1.55g}{1} \times \frac{1 \ mol}{46.1g} = 0.0336 \ mol }[/math]

Also, from the Thermal Energy equation, we can calculate the Heat transferred to the water:

[math]\displaystyle{ \Delta Q = mC \Delta T }[/math] (1)

The following are known values:

[math]\displaystyle{ m = 200g }[/math]

[math]\displaystyle{ C = 4.18 \ \frac{J}{g \cdot °C} }[/math]

[math]\displaystyle{ \Delta T = 55°C }[/math]

We can plug these values into 1 and get:

[math]\displaystyle{ \Delta Q_{water} = 200 \times 4.18 \times 55 = 45,980 \ J = 45.98 \ kJ }[/math]

By conservation of energy, we can reason that the Heat gained by the water sample is equal to the Heat lost by the Ethanol in the process of combustion. Hence, we make the following assumption:

[math]\displaystyle{ \Delta Q_{water} = Q_{Ethanol} }[/math], where

[math]\displaystyle{ Q_{Ethanol} = }[/math] the Heat of Combustion of this specific Ethanol sample

We can use this to find the Molar Heat of Combustion of Ethanol ([math]\displaystyle{ \boldsymbol{\Omega} }[/math]) as follows:

[math]\displaystyle{ \boldsymbol{\Omega} = \frac{Q_{Ethanol}}{M} = \frac{\Delta Q_{water}}{M} = \frac{45.98 \ kJ}{0.0336 \ mol} = 1,368.4 \ \frac{kJ}{mol} }[/math]

Middling

A bowl has a mass of 100 grams ([math]\displaystyle{ m }[/math]) while there is Ethanol in it. It is known from a previous measurement that the bowl had a mass of 50 grams ([math]\displaystyle{ m_{bowl} }[/math]).

A cup of water with mass 10 grams is heated from 20°C to 50°C by burning some of the sample of Ethanol.

a) Using these two statements, determine how much Ethanol must have combusted.

Let us call the mass of Ethanol combusted [math]\displaystyle{ m_{Ethanol_{c}} }[/math]. From statement one, we can figure out the total mass of the Ethanol sample, [math]\displaystyle{ m_{Ethanol_{t}} }[/math]:

[math]\displaystyle{ m = m_{bowl} + m_{Ethanol_{t}} }[/math]

Therefore:

[math]\displaystyle{ m_{Ethanol_{t}} = m - m_{bowl} = 100 - 50 = 50 }[/math] grams

Now, we want to figure out how much Heat must have been produced to warm the cup of water as much as was stated. We can do this with the Thermal Energy Equation:

[math]\displaystyle{ \Delta Q_{water} = m_{water}C_{water} \Delta T_{water} }[/math]

We know enough of the quantities in the equation to solve for the change in Thermal Energy:

[math]\displaystyle{ \Delta Q_{water} = 10 \times 4.18 \times (50 - 20) = 1254 \ J }[/math] (1)

Referring back to the Simple example, we know the Molar Heat of Combustion of Ethanol ([math]\displaystyle{ \Omega_{Ethanol} }[/math]):

[math]\displaystyle{ \Omega = 1,368.4 \ \frac{J}{mol} }[/math] (2)

We can find the number of moles of Ethanol ([math]\displaystyle{ M_{Ethanol} }[/math]) required to create the amount of Heat in 1, using the Molar Heat of Combustion of Ethanol:

[math]\displaystyle{ M_{Ethanol} = \Delta Q_{water} \frac{1}{\Omega_{Ethanol}} = 1254 \ \left(\frac{J}{1 }\right) \times \frac{1}{1,368.4} \ \left(\frac{mol}{J}\right) = 0.916 \ mol }[/math]

Using the molar mass of Ethanol [math]\displaystyle{ \left(Mm_{Ethanol} = 46.1 \ \frac{g}{mol}\right) }[/math], we can find the number of grams of Ethanol ([math]\displaystyle{ m_{ethanol_{c}} }[/math]) used to cause the necessary combustion:

[math]\displaystyle{ m_{Ethanol_{c}} = M_{Ethanol} \times Mm_{Ethanol} = 0.916 \ \left(\frac{mol}{1}\right) \times 46.1 \ \left(\frac{g}{mol}\right) = 42.23 }[/math] grams

We have found that 42.23 grams of the 50 gram Ethanol sample was combusted to Heat the water.

Difficult

What is the energy released when 1 kg of propane (C3H8) reacts with excess oxygen to form carbon dioxide and water? Bond enthalpies : C-H is 412 kJ/mol. C-C is 348 kJ mol, O=O is 498 kj/mol, H-O is 460 and C=O is 732

First, the chemical equation must be identified and balanced. 2C3H8 + 5O2 yields 3CO2 + 4H2O Then, the bond enthalpies of the reactants must be summed.

(2 C-C x 348 + 8 C-H x 412) x 2 = 7984.
1 O=O x 495 x 5 = 2475.
Their sum is 10459 kj/mol.
For the reactants.
2 C=O x 732 x 3 = 2988 kj/mol.
2 H-O x 460 x 4 = 3680.
Their sum is 8072 kj mol.
Subtracting according to Hess's law shows an enthalpy of -2,387 kj/mol.
We have a mass of propane which must be converted to mols. Since the molar mass of propane is 44.1 g/ mol, there are 22.67 mols of propane.
This reacts completely with excess oxygen.
22.67 mols x -2,387 kj/mol = -54,126.98 kJ energy.

Connectedness

Determining the heat of combustion of a fuel is essential when it is used an Energy source. Rockets, for example, convert chemical energy into gravitational potential energy and kinetic energy. Thus, knowing the necessary fuel to accomplish an escape speed without using too much space is critical. Chemical reactions are key as an energy source whether they provide electrical energy or thermal energy. My interest in health also contributes to enthalpy formation, specifically with the macromolecules our body prioritize to store energy. The bonds in carbohydrates do are more readily reactive and thus have a lower heat of formation than carbohydrates. Despite their composition have similar molecules, the bonding of each allows for a larger heat of formation when fats are consumed than when carbohydrates are, leading them to be stored as they are a better source of energy for less space.

History

In 1840, Russian chemist Germain Hess proposed that the summation of enthalpy between the initial reactants of a multi-step reaction and the final products was the same no matter the pathways taken. He proposed that the change in enthalpy was a state function, and thus did not depend on the pathway taken so long as the overall reactants and products are the same. He experimented on the hydrates of sulfuric acids and determined that the heat released was the same no matter the path taken. He was a primarily experimental chemist and was concerned with how the affinity of molecules impacts its release in thermal energy.

See also

Further reading

• Wikipedia: Heat of Combustion
  
• CK-12: Heat of Combustion
  
• CK-12: Heat of Combustion 2
  
• The Mole in Chemical Reactions
  

External links

• Kinds of Matter
  
• The Energy Principle
  
• Conservation of Energy
  
• System & Surroundings
  
• Thermal Energy
  
• Specific Heat
  
• Temperature
  

References

• Wikipedia: Heat of Combustion
  
• CK-12: Heat of Combustion
  
• CK-12:Heat of Combustion
  
• The Mole in Chemical Reactions
  
• [1]