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Chemical transformations · Chemical thermodynamics

Enthalpy and calorimetry

At a football game, you crack an instant cold pack and it turns icy in seconds. Squeeze a hand warmer and it heats up. Both are just chemical reactions, so why does one take in heat while the other gives it off? And how can we put a number on that heat?

Enthalpy: heat at constant pressure

Most reactions happen in open beakers or in the air, at constant atmospheric pressure. Chemists have a quantity for the heat exchanged under those conditions.

The sign tells you the direction of the heat flow, always from the point of view of the chemicals (the system).

Two enthalpy diagrams side by side: in an exothermic reaction the products sit below the reactants, and in an endothermic reaction the products sit above the reactants. H reactants products ΔH < 0 exothermic reactants products ΔH > 0 endothermic

Try the widget. Change the energy values and watch where the products end up compared with the reactants.

Why heat is released or absorbed

Heat comes from the bonds. Atoms are held together by chemical bonds, and changing bonds changes energy.

  • Breaking bonds always absorbs energy. You have to pull the atoms apart.

  • Forming bonds always releases energy. The atoms settle into a lower-energy arrangement.

A reaction does both. If the new bonds formed in the products release more energy than it took to break the old bonds, the net result is heat released and the reaction is exothermic. If it takes more energy to break the old bonds than the new bonds give back, the reaction is endothermic.

Energy diagram of a reaction: breaking bonds raises the energy, forming new bonds lowers it, and the net change is the enthalpy change. energy reactants separated atoms products break bonds: absorbs energy form bonds: releases energy

Here the energy released forming bonds is larger than the energy absorbed breaking bonds, so this reaction is exothermic.

Thermochemical equations

$$\mathrm{2\,H_2(g) + O_2(g) \rightarrow 2\,H_2O(l)} \qquad \Delta H^\circ = -571.6\ \text{kJ}$$

This means 571.6 kJ is released when 2 mol of $\mathrm{H_2}$ react with 1 mol of $\mathrm{O_2}$ to form 2 mol of liquid water.

Practice★☆☆Not started
$A 298\,\text{K}$, 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:

$$2\,\mathrm{CO(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{CO_2(g)} \quad \Delta H_1^\circ = -566.0\,\text{kJ}$$
  1. a) Determine the value of the standard enthalpy of the oxidation reaction of carbon monoxide, expressed per mole of $\mathrm{CO_2(g)}$ formed.
  2. b) Calculate the standard enthalpy change, $\Delta H_2^\circ$, for the decomposition reaction:
$$\mathrm{CO_2(g)} \rightarrow \mathrm{CO(g)} + \frac{1}{2}\,\mathrm{O_2(g)}$$
Solve
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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:

H (kJ) Reaction progress 0 kJ −92 kJ N₂(g) + 3 H₂(g) 2 NH₃(g) ? ΔH

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.

  1. 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.
  2. 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: $M(\mathrm{NH_3}) = 17.0\,\text{g}\times\text{mol}^{-1}$.

Solve

From grams of reactant to kilojoules

When you burn the propane in a grill tank, how much heat do you get? The heat released or absorbed is proportional to how much reacts.

Flow chart: divide the mass in grams by the molar mass to get moles, then multiply by the enthalpy change per mole to get the heat in kilojoules. mass (g) moles (mol) heat (kJ) ÷ M × ΔH
Practice★☆☆Not started
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 $1\,\text{bar}$:

$$\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)} \quad \Delta H^\circ = +178.0\,\text{kJ}\times\text{mol}^{-1}$$
  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.
  2. b) Calculate the heat exchanged, expressed in kilojoules, when $25.0\,\text{g}$ of pure calcium carbonate decompose completely.

Data: Atomic molar masses: $M(\mathrm{Ca}) = 40.1\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{C}) = 12.0\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{O}) = 16.0\,\text{g}\times\text{mol}^{-1}$.

Solve
Practice★☆☆Not started
Two fuels are available for home heating: methane, $\mathrm{CH_4(g)}$, and propane, $\mathrm{C_3H_8(g)}$. The standard thermochemical equations for complete combustion…

Two fuels are available for home 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:

$$\mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)} \quad \Delta H^\circ = -890.4\,\text{kJ}\times\text{mol}^{-1}$$ $$\mathrm{C_3H_8(g)} + 5\,\mathrm{O_2(g)} \rightarrow 3\,\mathrm{CO_2(g)} + 4\,\mathrm{H_2O(l)} \quad \Delta H^\circ = -2220.0\,\text{kJ}\times\text{mol}^{-1}$$
  1. a) Calculate the energy released per gram of fuel consumed in each case.
  2. b) State, justifying your answer with the results obtained, which of the two hydrocarbons provides the greater energy yield per unit of mass.

Data: Atomic molar masses: $M(\mathrm{C}) = 12.0\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{H}) = 1.0\,\text{g}\times\text{mol}^{-1}$.

Solve

Measuring heat: the coffee-cup calorimeter

How do we find $\Delta H$ in the lab? We let the reaction happen in water inside a coffee-cup calorimeter: an insulated foam cup open to the air, so the pressure is constant. The heat released by the reaction is absorbed by the solution, and a thermometer measures the temperature change.

Practice★★☆Not started
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 neutralization is complete, the final temperature of the homogeneous mixture stabilizes at $24.4\,^\circ\text{C}$.

  1. a) Write the net ionic equation of the process and determine the amount of substance, in moles, of water formed.
  2. b) Calculate the experimental molar enthalpy of neutralization, 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.

Solve

Combining equations and enthalpies of formation

Enthalpy depends only on the starting and ending states, not on the path. So if you add chemical equations, you add their $\Delta H$ values.

The standard enthalpy of formation, $\Delta H_f^\circ$, is the enthalpy change for forming 1 mol of a compound from its elements in their standard states. For an element in its standard state (such as $\mathrm{O_2(g)}$ or $\mathrm{C(s,\ graphite)}$), $\Delta H_f^\circ = 0$.

Practice★★★Not started
Gaseous hydrogen, $\mathrm{H_2(g)}$, and liquid hydrazine, $\mathrm{N_2H_4(l)}$, are compared as aerospace fuels. Under standard conditions at $298\,\text{K}$, the molar…

Gaseous hydrogen, $\mathrm{H_2(g)}$, and liquid hydrazine, $\mathrm{N_2H_4(l)}$, are compared as aerospace fuels. Under standard conditions at $298\,\text{K}$, the molar enthalpies of formation are $\Delta H_f^\circ[\mathrm{H_2O(l)}] = -285.8\,\text{kJ}\times\text{mol}^{-1}$ and $\Delta H_f^\circ[\mathrm{N_2H_4(l)}] = +50.6\,\text{kJ}\times\text{mol}^{-1}$.

The complete combustion reactions with oxygen gas form, respectively, liquid water in the case of hydrogen, and nitrogen gas together with liquid water in the case of hydrazine.

  1. a) Write both balanced thermochemical equations per mole of fuel consumed and determine the standard enthalpy of combustion of each in $\text{kJ}\times\text{mol}^{-1}$.
  2. b) If a rocket has a tank with a payload limit of $500\,\text{kg}$ of fuel, work out, with justification, which substance will supply the greater total energy and determine the volume of dry gaseous product that would remain after cooling and completely condensing the water (at $1\,\text{bar}$ and $298\,\text{K}$) for each fuel. Consider only the stoichiometric products: just enough oxygen, with no air or inert gas. This cold volume does not represent the hot exhaust of an engine.

Data: $M(\mathrm{H}) = 1.0\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{N}) = 14.0\,\text{g}\times\text{mol}^{-1}$, $R = 0.0831\,\text{bar}\times\text{L}\times\text{mol}^{-1}\times\text{K}^{-1}$.

Solve
Practice★★★Not started
In a lab experiment to determine the enthalpy of hydration and subsequent neutralization of calcium oxide, a student carries out two experiments in a constant-pressure…

In a lab experiment to determine the enthalpy of hydration and subsequent neutralization of calcium oxide, a student carries out two experiments in a constant-pressure calorimeter with negligible heat capacity:

  1. $2.80\,\text{g}$ of $\mathrm{CaO(s)}$ are reacted in $200.0\,\text{g}$ of liquid water, producing the reaction $\mathrm{CaO(s)} + \mathrm{H_2O(l)} \rightarrow \mathrm{Ca(OH)_2(s)}$, and the water is observed to heat up, absorbing $3.26\,\text{kJ}$.

  2. $3.70\,\text{g}$ of $\mathrm{Ca(OH)_2(s)}$ are reacted with an excess of aqueous solution of $\mathrm{HCl}$, releasing $5.84\,\text{kJ}$ according to: $\mathrm{Ca(OH)_2(s)} + 2\,\mathrm{HCl(aq)} \rightarrow \mathrm{CaCl_2(aq)} + 2\,\mathrm{H_2O(l)}$.

  1. a) Calculate the standard molar enthalpy change for each of the two reactions above.
  2. b) Applying Hess's law, determine the standard enthalpy for the direct reaction:
$$\mathrm{CaO(s)} + 2\,\mathrm{HCl(aq)} \rightarrow \mathrm{CaCl_2(aq)} + \mathrm{H_2O(l)}$$

Data: Atomic molar masses: $M(\mathrm{Ca}) = 40.1\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{O}) = 16.0\,\text{g}\times\text{mol}^{-1}$, $M(\mathrm{H}) = 1.0\,\text{g}\times\text{mol}^{-1}$. Consider the heats corrected to 298 K and standard pressure; neglect dilution effects and heat losses.

Solve

ΔH versus ΔU

Enthalpy is defined as $H = U + PV$, where $U$ is internal energy. At constant pressure, a reaction that makes or uses gas pushes the atmosphere back or lets it in, which does expansion work. That is why $\Delta H$ and $\Delta U$ differ slightly: $$\Delta H = \Delta U + \Delta n_g\,RT$$ Here $\Delta n_g$ is the moles of gaseous products minus the moles of gaseous reactants, $R = 8.314$ J/(mol·K), and $T$ is in kelvins. Only gases count, since solids and liquids barely change volume.

Back to the cold pack and hand warmer: the cold pack's dissolving process absorbs heat from its surroundings, so $\Delta H > 0$, while the hand warmer's reaction releases heat, so $\Delta H < 0$. You can now measure those values with a foam cup and a thermometer.

Summary

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