Enthalpy and calorimetry
After a hurling match, a hand-warmer in your pocket heats up as soon as you open the packet, while a cold pack (instant ice) on a sore ankle gets freezing cold when you squeeze it. Both are chemical reactions in a plastic bag. Why does one release heat and the other absorb it, and how can we measure and predict how much?
Enthalpy change and heat of reaction
Most reactions in the lab happen in open containers at atmospheric pressure. The heat exchanged under these conditions has its own name.
We always take the point of view of the reaction itself:
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Exothermic: heat is given out to the surroundings, so $\Delta H < 0$. The surroundings (for example the solution in a beaker) get warmer.
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Endothermic: heat is taken in from the surroundings, so $\Delta H > 0$. The surroundings get colder.
Examples from the course: combustion, neutralisation, rusting of iron and respiration are exothermic. Thermal decomposition of calcium carbonate and photosynthesis are endothermic. Changes of state also involve enthalpy changes: melting and evaporating are endothermic, freezing and condensing are exothermic.
Try the interactive profile and change $\Delta H$ to see which way the diagram moves.
Calcium carbonate decomposes thermally according to the following thermochemical equation at $298\,\text{K}$ and $1\,\text{bar}$: $$\mathrm{CaCO_3(s)} \rightarrow…
Calcium carbonate decomposes thermally according to the following thermochemical equation at $298\,\text{K}$ and $1\,\text{bar}$:
$$\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)} \quad \Delta H^\circ = +178.0\,\text{kJ}\times\text{mol}^{-1}$$- a) Justify whether the process is exothermic or endothermic and which physical quantity this enthalpy change represents if the reaction is carried out in an open container at constant pressure.
- b) Calculate the heat exchanged, expressed in kilojoules, when $25.0\,\text{g}$ of pure calcium carbonate decompose completely.
Data: Atomic molar masses: $M(\mathrm{Ca}) = 40.1\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{C}) = 12.0\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{O}) = 16.0\,\text{g}\times\text{mol}^{-1}$.
Why reactions give out or take in heat
In any reaction, bonds in the reactants break (energy in) and new bonds form in the products (energy out).
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If more energy is released making bonds than was needed to break bonds, the reaction is exothermic.
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If more energy is needed to break bonds than is released, it is endothermic.
This is conservation of energy: the difference shows up as heat exchanged with the surroundings.
You can measure $\Delta H$ for a neutralisation in a school practical. Place a polystyrene cup in a beaker for support and add a measured volume of acid. Record its temperature with a thermometer for a few minutes. Add the same volume of base at the same temperature, stir gently with the thermometer, and record the highest temperature reached. The temperature rise is $\Delta T$, and you use it in the calculation shown in the calorimetry section below.
Thermochemical equations
If you multiply the coefficients, you multiply $\Delta H$. If you reverse the equation, you change the sign of $\Delta H$.
Energy profile diagrams can be drawn directly from a thermochemical equation. $\Delta H = H(\text{products}) - H(\text{reactants})$, so it is the vertical gap between the two levels. For the Haber process, $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$, $\Delta H = -92.0\ \text{kJ}$: the reactants ($\text{N}_2 + 3\text{H}_2$) are at $0\ \text{kJ}$ and the products ($2\text{NH}_3$) at $-92.0\ \text{kJ}$, so the products sit $92.0\ \text{kJ}$ below the reactants and the arrow for $\Delta H$ points down. The curve rises to a peak above both levels: this is the activation energy barrier.
Standard heat of formation
The standard heat of formation, $\Delta H_f^\circ$, is the enthalpy change when one mole of a compound forms from its elements in their standard states. For an element in its standard state, $\Delta H_f^\circ = 0$. You can then find $\Delta H$ for a reaction from these values: $$\Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$$ Multiply each value by its coefficient in the balanced equation.
$A 298\,\text{K}$, the combustion of carbon monoxide gas to give carbon dioxide proceeds with release of energy according to: $$2\,\mathrm{CO(g)} + \mathrm{O_2(g)}…
$A 298\,\text{K}$, the combustion of carbon monoxide gas to give carbon dioxide proceeds with release of energy according to:
$$2\,\mathrm{CO(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{CO_2(g)} \quad \Delta H_1^\circ = -566.0\,\text{kJ}$$- a) Determine the value of the standard enthalpy of the oxidation reaction of carbon monoxide, expressed per mole of $\mathrm{CO_2(g)}$ formed.
- b) Calculate the standard enthalpy change, $\Delta H_2^\circ$, for the decomposition reaction:
The industrial synthesis of ammonia is described by the Haber-Bosch process. The following enthalpy diagram shows the energy associated with this thermochemical…
The industrial synthesis of ammonia is described by the Haber-Bosch process. The following enthalpy diagram shows the energy associated with this thermochemical transformation at $298\,\text{K}$:
The figure outlines the enthalpy balance of the reaction:
The gaseous reactants are at a relative enthalpy of $0\,\text{kJ}$ and the products at $-92.0\,\text{kJ}$ per mole of $\mathrm{N_2(g)}$ that reacts.
- a) Write the balanced thermochemical equation with the smallest integer stoichiometric coefficients and deduce, with justification, the sign and numerical value indicated by the arrow with the question mark.
- b) Calculate the amount of heat exchanged at constant pressure when $0.50\,\text{mol}$ of $\mathrm{N_2(g)}$ react completely with excess hydrogen, and determine the heat required to decompose $68.0\,\text{g}$ of gaseous ammonia into its elements at the same temperature.
Data: Molar mass: $M(\mathrm{NH_3}) = 17.0\,\text{g}\times\text{mol}^{-1}$.
From mass to heat
Calorimetry
In the lab, $\Delta H$ is found by measuring the temperature change of water or a solution in a insulated calorimeter (a polystyrene cup). The heat gained by the solution is $$q = mc\Delta T$$ where $m$ is the mass of solution, $c$ the specific heat capacity ($4.18\ \text{J g}^{-1}\text{K}^{-1}$ for water or dilute solutions) and $\Delta T$ the temperature rise. The reaction gave out this heat, so if the solution gets warmer, $\Delta H$ is negative.
In the spirit-burner experiment you burn an alcohol under a can of water and use $q = mc\Delta T$ the same way. The experimental value is usually smaller in magnitude than the standard value, because heat is lost to the air and the can, and combustion may be incomplete.
Two fuels are available for domestic heating: methane, $\mathrm{CH_4(g)}$, and propane, $\mathrm{C_3H_8(g)}$. The standard thermochemical equations for complete…
Two fuels are available for domestic heating: methane, $\mathrm{CH_4(g)}$, and propane, $\mathrm{C_3H_8(g)}$. The standard thermochemical equations for complete combustion at $298\,\text{K}$ are:
$$\mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)} \quad \Delta H^\circ = -890.4\,\text{kJ}\times\text{mol}^{-1}$$ $$\mathrm{C_3H_8(g)} + 5\,\mathrm{O_2(g)} \rightarrow 3\,\mathrm{CO_2(g)} + 4\,\mathrm{H_2O(l)} \quad \Delta H^\circ = -2220.0\,\text{kJ}\times\text{mol}^{-1}$$- a) Calculate the energy released per gram of fuel consumed in each case.
- b) State, justifying your answer with the results obtained, which of the two hydrocarbons provides the greater energy yield per unit mass.
Data: Atomic molar masses: $M(\mathrm{C}) = 12.0\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{H}) = 1.0\,\text{g}\times\text{mol}^{-1}$.
In a constant-pressure calorimeter of $1\,\text{bar}$ (whose heat capacity is considered negligible), $100.0\,\text{mL}$ of an aqueous solution of…
In a constant-pressure calorimeter of $1\,\text{bar}$ (whose heat capacity is considered negligible), $100.0\,\text{mL}$ of an aqueous solution of $\mathrm{HCl} 0.50\,\text{M}$ are mixed with $100.0\,\text{mL}$ of an aqueous solution of $\mathrm{NaOH} 0.50\,\text{M}$, both initially at $21.0\,^\circ\text{C}$. After neutralisation is complete, the final temperature of the homogeneous mixture stabilises at $24.4\,^\circ\text{C}$.
- a) Write the net ionic equation of the process and determine the amount of substance, in moles, of water formed.
- b) Calculate the experimental molar enthalpy of neutralisation, expressed in $\text{kJ}\times\text{mol}^{-1}$ of water produced.
Data: Assume that the density of all aqueous solutions is $1.00\,\text{g}\times\text{mL}^{-1}$ and that their mean specific heat capacity is equal to that of pure water, ce $= 4.18\,\text{J}\times\text{g}^{-1}\times\text{K}^{-1}$. Assume there are no heat losses to the surroundings.
Enthalpy change and internal energy change
When gases are made or used up, the reaction pushes against the atmosphere, so the heat at constant pressure ($\Delta H$) differs slightly from the change in internal energy ($\Delta U$), the heat at constant volume, as measured in a bomb calorimeter.
Bomb calorimeter
In a bomb calorimeter the reaction happens at constant volume inside a steel bomb surrounded by water. The bomb absorbs heat too, so you include its heat capacity $C$ (in $\text{J K}^{-1}$): $$q = (mc + C)\Delta T$$ where $m$ is the mass of water. This heat equals $-\Delta U$ for the amount burned.
$A 298\,\text{K}$ and $1\,\text{bar}$, the complete combustion of liquid ethanol experimentally released heat at constant pressure of…
$A 298\,\text{K}$ and $1\,\text{bar}$, the complete combustion of liquid ethanol experimentally released heat at constant pressure of $1366.8\,\text{kJ}\times\text{mol}^{-1}$ according to:
$$\mathrm{C_2H_5OH(l)} + 3\,\mathrm{O_2(g)} \rightarrow 2\,\mathrm{CO_2(g)} + 3\,\mathrm{H_2O(l)}$$- a) Deduce the general formal relationship between the enthalpy change $(\Delta H)$ and the change in internal energy $(\Delta U)$ for an ideal gas-phase reaction and calculate the value of $\Delta n_g$ for this reaction.
- b) Calculate the standard molar change in internal energy, $\Delta U^\circ$, in $\text{kJ}\times\text{mol}^{-1}$, for the combustion of ethanol at that temperature.
Data: Universal ideal gas constant: $R = 8,314\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$. Assume ideal behaviour for gaseous substances and negligible volume of liquids.
A sample of $1,150\,\text{g}$ of solid benzoic acid, $\mathrm{C_6H_5COOH(s)}$, is completely burned inside a constant-volume bomb calorimeter surrounded by…
A sample of $1,150\,\text{g}$ of solid benzoic acid, $\mathrm{C_6H_5COOH(s)}$, is completely burned inside a constant-volume bomb calorimeter surrounded by $1200\,\text{g}$ of pure water. The heat capacity of the bomb itself (dry) is $C_b = 850.0\,\text{J}\times\text{K}^{-1}$. During complete combustion, the temperature of the assembly rises from $22.15\,^\circ\text{C}$ to $27.35\,^\circ\text{C}$.
- a) Write the thermochemical equation for the complete combustion of benzoic acid producing gaseous carbon dioxide and liquid water, and calculate the molar change in internal energy per mole of acid burned $(\Delta U^\circ)$.
- b) Calculate the standard enthalpy change of combustion, $\Delta H^\circ$, in $\text{kJ}\times\text{mol}^{-1}$, at $298\,\text{K}$.
Data: Atomic molar masses: $M(\mathrm{C}) = 12.0\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{H}) = 1.0\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{O}) = 16.0\,\text{g}\times\text{mol}^{-1}$. ce$(\mathrm{agua}) = 4.18\,\text{J}\times\text{g}^{-1}\times\text{K}^{-1}$. $R = 8,314\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$. Assume the correction for the small temperature change relative to 298 K and the effect of pressure on the properties of the solid are negligible.
Back to the start: the hand-warmer is exothermic ($\Delta H < 0$, more energy released making bonds than absorbed breaking them), and the cold pack is endothermic ($\Delta H > 0$).