Hess's Law and Bond Energies
Imagine trying to measure how much heat is released when graphite and hydrogen gas combine to form acetylene (ethyne), the fuel in a welding torch. In a school lab, that reaction is nearly impossible to carry out cleanly. Hess's law gets around the problem: you combine reactions that are easy to measure and calculate the one you can't.
Hess's Law
Burn 1 mol of carbon with plenty of oxygen in a calorimeter and 393.5 kJ is released as $\mathrm{CO_2(g)}$. Now burn it in two steps, first to $\mathrm{CO(g)}$ and then to $\mathrm{CO_2(g)}$. The heat released in the second step, 283.0 kJ, is known, but the first step is hard to measure on its own. Do the two routes still have to release the same total? Why wouldn't the total heat depend on the path?
At $298\,\mathrm{K}$, the conversion of sulfur dioxide into sulfur trioxide, a central step in the industrial synthesis of sulfuric acid, is to be evaluated:…
At $298\,\mathrm{K}$, the conversion of sulfur dioxide into sulfur trioxide, a central step in the industrial synthesis of sulfuric acid, is to be evaluated:
$$\mathrm{SO}_2(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_3(\mathrm{g})$$Experimental data are available for the following combustion and oxidation reactions at the same temperature and pressure:
(1) $\mathrm{S}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_2(\mathrm{g}) \quad \Delta H_1^\circ = -296.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$
(2) $\mathrm{S}(\mathrm{s}) + \frac{3}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_3(\mathrm{g}) \quad \Delta H_2^\circ = -395.7\,\mathrm{kJ}\times\mathrm{mol}^{-1}$
- a) Using Hess's law, combine equations (1) and (2) algebraically to deduce the standard enthalpy of the target reaction.
- b) Explain briefly which property of the state function enthalpy Hess's law is based on.
The combustion of graphite carbon can take place completely to give carbon dioxide or in stages via the intermediate carbon monoxide. The energy relationships between…
The combustion of graphite carbon can take place completely to give carbon dioxide or in stages via the intermediate carbon monoxide. The energy relationships between these transformations at $298\,\mathrm{K}$ are summarized in the following triangular diagram:
The thermochemical data at $298\,\mathrm{K}$ and $1\,\mathrm{bar}$ are:
(Direct route) $\mathrm{C}(\text{grafito}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \quad \Delta H_1^\circ = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$
(Second stage) $\mathrm{CO}(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \quad \Delta H_3^\circ = -283.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$
- a) Determine the value of the standard enthalpy of formation of gaseous carbon monoxide, $\Delta H_2^\circ = \Delta H_\mathrm{f}^\circ[\mathrm{CO}(\mathrm{g})]$, corresponding to the first stage.
- b) Justify in detail the path independence, using the cycle in the diagram.
Enthalpies of Formation
Millions of reactions are known, and nobody could measure each one. Is there a single reference value for each compound that lets you predict the enthalpy of any reaction without running it?
At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$, the thermal decomposition reaction of calcium carbonate for the industrial production of quicklime is studied:…
At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$, the thermal decomposition reaction of calcium carbonate for the industrial production of quicklime is studied:
$$\mathrm{CaCO}_3(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})$$- a) Calculate the standard enthalpy of this reaction from the standard molar enthalpies of formation: $\Delta H_\mathrm{f}^\circ[\mathrm{CaCO}_3(\mathrm{s})] = -1206.9\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $\Delta H_\mathrm{f}^\circ[\mathrm{CaO}(\mathrm{s})] = -635.1\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $\Delta H_\mathrm{f}^\circ[\mathrm{CO}_2(\mathrm{g})] = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
- b) Justify the thermodynamic sign obtained for the process and state, with justification, what the enthalpy of formation is for any chemical element in its most stable reference form at that temperature and at standard pressure.
At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$, samples of graphite carbon, hydrogen gas and acetylene gas (ethyne, $\mathrm{C}2\mathrm{H}2$) are completely burned in an…
At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$, samples of graphite carbon, hydrogen gas and acetylene gas (ethyne, $\mathrm{C}_2\mathrm{H}_2$) are completely burned in an oxygen atmosphere, and the following standard enthalpies of combustion are measured:
(1) $\mathrm{C}(\text{grafito}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) \quad \Delta H_\mathrm{c,1}^\circ = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$
(2) $\mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{l}) \quad \Delta H_\mathrm{c,2}^\circ = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$
(3) $\mathrm{C}_2\mathrm{H}_2(\mathrm{g}) + \frac{5}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow 2\,\mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\mathrm{l}) \quad \Delta H_\mathrm{c,3}^\circ = -1299.6\,\mathrm{kJ}\times\mathrm{mol}^{-1}$
- a) Applying Hess's law, calculate the standard molar enthalpy of formation of gaseous acetylene from its elements in their standard states: $2\,\mathrm{C}(\text{grafito}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_2\mathrm{H}_2(\mathrm{g})$.
- b) Calculate the heat exchanged at constant pressure, stating whether it is absorbed or released, when $52.0\,\mathrm{g}$ of acetylene are formed under these conditions. (Molar mass of $\mathrm{C}_2\mathrm{H}_2 = 26.0\,\mathrm{g}\times\mathrm{mol}^{-1}$).
The direct synthesis of ethanol from its elements under standard conditions,…
The direct synthesis of ethanol from its elements under standard conditions, $2\,\mathrm{C}(\text{grafito}) + 3\,\mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}(\mathrm{l})$, is experimentally unfeasible directly with quantitative purity owing to the simultaneous formation of multiple hydrocarbons and oxygenated compounds.
- a) Design a Hess cycle that allows $\Delta H_\mathrm{f}^\circ[\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}(\mathrm{l})]$ to be calculated at $298\,\mathrm{K}$ given that the standard molar enthalpies of combustion are:
$\Delta H_\mathrm{c}^\circ[\mathrm{C}(\text{grafito})] = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $\Delta H_\mathrm{c}^\circ[\mathrm{H}_2(\mathrm{g})] = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $\Delta H_\mathrm{c}^\circ[\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}(\mathrm{l})] = -1366.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
- b) Determine the standard enthalpy of formation of ethanol in the gas phase given that the enthalpy of vaporization of liquid ethanol at $298\,\mathrm{K}$ is $\Delta H_\mathrm{vap}^\circ[\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}(\mathrm{l})] = +42.3\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, and interpret the effect of intermolecular condensation on the enthalpy difference between liquid and gas (do not deduce spontaneity from that datum alone).
Bond Energies: A Gas-Phase Estimate
Mix hydrogen and chlorine gas and a spark of light sets off a violent, strongly exothermic reaction that forms hydrogen chloride gas. Only covalent bonds are involved. Where does all that energy come from?
The catalytic hydrogenation of ethene in the gas phase leads to the formation of ethane according to the following thermochemical equation at $298\,\mathrm{K}$:…
The catalytic hydrogenation of ethene in the gas phase leads to the formation of ethane according to the following thermochemical equation at $298\,\mathrm{K}$:
$$\mathrm{CH}_2=\mathrm{CH}_2(\mathrm{g}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{CH}_3-\mathrm{CH}_3(\mathrm{g})$$- a) Estimate the standard enthalpy change of the process using only the mean bond energy values: $E(\mathrm{C}=\mathrm{C}) = 611.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{C}-\mathrm{C}) = 347.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{C}-\mathrm{H}) = 414.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $E(\mathrm{H}-\mathrm{H}) = 436.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
- b) Explain why mean bond energies allow an estimate in the gas phase, not an exact value, and what corrections would have to be added if any species were condensed.
The aim is to estimate the standard enthalpy of formation of liquid water, $\Delta H_\mathrm{f}^\circ[\mathrm{H}2\mathrm{O}(\mathrm{l})]$, at $298\,\mathrm{K}$ using…
The aim is to estimate the standard enthalpy of formation of liquid water, $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})]$, at $298\,\mathrm{K}$ using mean bond energies, for which the following are available:
$E(\mathrm{H}-\mathrm{H}) = 436.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{O}=\mathrm{O}) = 498.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{O}-\mathrm{H}) = 463.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, and the enthalpy of vaporization of water at $298\,\mathrm{K}$: $\Delta H_\mathrm{vap}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})] = +44.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
- a) Calculate the standard enthalpy of the gas-phase reaction $\mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\,\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{g})$ from the bond energies, and from it deduce $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})]$ by including the change of state.
- b) Compare the result with the experimental literature value of $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})] = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and state two factors that cause the mean bond energy method to introduce deviations.
For the combustion of propane gas at $298\,\mathrm{K}$: $$\mathrm{CH}3\mathrm{CH}2\mathrm{CH}3(\mathrm{g}) + 5\,\mathrm{O}2(\mathrm{g}) \rightarrow…
For the combustion of propane gas at $298\,\mathrm{K}$:
$$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_3(\mathrm{g}) + 5\,\mathrm{O}_2(\mathrm{g}) \rightarrow 3\,\mathrm{CO}_2(\mathrm{g}) + 4\,\mathrm{H}_2\mathrm{O}(\mathrm{g})$$- a) Calculate the standard enthalpy change of this reaction using only the average bond energy data: $E(\mathrm{C}-\mathrm{C}) = 347.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{C}-\mathrm{H}) = 414.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{O}=\mathrm{O}) = 498.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $E(\mathrm{C}=\mathrm{O}) = 799.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ (in $\mathrm{CO}_2$) and $E(\mathrm{O}-\mathrm{H}) = 463.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
- b) Calculate the enthalpy of combustion from the standard molar enthalpies of formation: $\Delta H_\mathrm{f}^\circ[\mathrm{C}_3\mathrm{H}_8(\mathrm{g})] = -103.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, $\Delta H_\mathrm{f}^\circ[\mathrm{CO}_2(\mathrm{g})] = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{g})] = -241.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$. Discuss the physical origin of the quantitative discrepancy observed between the two analytical methods.
In a calorimetric laboratory, the standard enthalpy of combustion of the solvent cyclohexane, $\mathrm{C}6\mathrm{H}{12}(\mathrm{l})$, is evaluated. The following…
In a calorimetric laboratory, the standard enthalpy of combustion of the solvent cyclohexane, $\mathrm{C}_6\mathrm{H}_{12}(\mathrm{l})$, is evaluated. The following reaction data at $298\,\mathrm{K}$ are obtained by combustion at constant volume and correction to constant pressure:
(1) $\mathrm{C}_6\mathrm{H}_{12}(\mathrm{l}) + 9\,\mathrm{O}_2(\mathrm{g}) \rightarrow 6\,\mathrm{CO}_2(\mathrm{g}) + 6\,\mathrm{H}_2\mathrm{O}(\mathrm{l}) \quad \Delta H_1^\circ = -3919.6\,\mathrm{kJ}\times\mathrm{mol}^{-1}$
The standard enthalpies of formation of $\mathrm{CO}_2(\mathrm{g})$ and of liquid water are also available: $\Delta H_\mathrm{f}^\circ[\mathrm{CO}_2(\mathrm{g})] = -393.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and $\Delta H_\mathrm{f}^\circ[\mathrm{H}_2\mathrm{O}(\mathrm{l})] = -285.8\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, as well as the enthalpy of vaporization of cyclohexane: $\Delta H_\mathrm{vap}^\circ[\mathrm{C}_6\mathrm{H}_{12}(\mathrm{l})] = +33.1\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
- a) Determine the standard enthalpy of formation of liquid cyclohexane, $\Delta H_\mathrm{f}^\circ[\mathrm{C}_6\mathrm{H}_{12}(\mathrm{l})]$, and of gaseous cyclohexane, $\Delta H_\mathrm{f}^\circ[\mathrm{C}_6\mathrm{H}_{12}(\mathrm{g})]$.
- b) If the experimental combustion in the calorimeter had been incomplete, forming a molar $1.0\,\%$ of $\mathrm{CO}(\mathrm{g})$ (whose $\Delta H_\mathrm{f}^\circ[\mathrm{CO}(\mathrm{g})] = -110.5\,\mathrm{kJ}\times\mathrm{mol}^{-1}$) instead of $\mathrm{CO}_2(\mathrm{g})$, explain whether the heat measured experimentally would be an overestimate or an underestimate relative to the theoretical stoichiometric value.
Physical states and constant-volume data
Check the state symbol of water. $\mathrm{H_2O(g)}$ and $\mathrm{H_2O(l)}$ differ by the enthalpy of vaporization: $\mathrm{H_2O(l)} \rightarrow \mathrm{H_2O(g)}$, $\Delta H_{vap}^\circ = +44.0\ \text{kJ/mol}$. Bond energies refer to gases, so they give the enthalpy for $\mathrm{H_2O(g)}$. To get liquid water, add $-44.0\ \text{kJ}$ for each mole condensed.
A bomb calorimeter measures heat at constant volume, which is very close to the value at constant pressure for most reactions. Labs apply a small correction, and the corrected value is the $\Delta H^\circ$ you use in Hess cycles. When a problem gives the corrected $\Delta H^\circ$, just treat it like any other equation.
Back to the start: To get the enthalpy of formation of acetylene, $2\,\mathrm{C(graphite)} + \mathrm{H_2(g)} \rightarrow \mathrm{C_2H_2(g)}$, combine the known combustion enthalpies of graphite ($-393.5\ \text{kJ/mol}$), hydrogen ($-285.8\ \text{kJ/mol}$) and acetylene ($-1299.5\ \text{kJ/mol}$): $\Delta H_f^\circ = 2(-393.5) + (-285.8) - (-1299.5) = +226.7\ \text{kJ/mol}$. No impossible direct synthesis needed.