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General Chemistry/Enthalpy and Bond Strength/Enthalpies of Reaction

Enthalpy and Bond Strength

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Enthalpies of Reaction

The enthalpy of a reaction is equal to the total enthalpy of products minus the total enthalpy of the reactants: ΔH=Hproducts−Hreactants\Delta H=H_{\rm{products}}-H_{\rm{reactants}}.
The enthalpy change that occurs during a reaction is called the enthalpy, or heat, of reaction. Reaction enthalpies of common reactions have been experimentally measured under constant pressure. Consider the reaction between hydrogen and oxygen to make water:
2H2(g)+O2(g)→2H2O(g)2{\rm H}_2(g)+{\rm O}_2(g)\rightarrow2{\rm H}_2{\rm O}(g)
When 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water, this reaction releases 483.6 kJ of heat. The change in enthalpy is equal to the enthalpy of the products minus the enthalpy of the reactants:
ΔH=Hproducts−Hreactants\Delta H=H_{\rm{products}}-H_{\rm{reactants}}
This reaction releases heat. This means the products have less energy than the reactants, and ΔH\Delta H will be a negative value. ΔH\Delta H is –483.6 kJ for this reaction. This sign convention holds true for all reactions: If a reaction releases heat (exothermic), then it has a negative ΔH\Delta H. If a reaction absorbs heat (endothermic), then ΔH\Delta H is positive. Reaction enthalpies are proportional to the masses of the substances involved. Enthalpy is an extensive property, a property that is determined by the amount of matter in a system. If 4 moles of hydrogen react with 2 moles of oxygen to produce 4 moles of water, the energy released would be
483.6×2=967.2kJ483.6\times2=967.2\;{\rm{kJ}}
If the chemical equation is reversed, the sign of the enthalpy of reaction is reversed too. Water can be broken down into its component elements through electrolysis.
2H2O(g)→2H2(g)+O2(g)ΔH=483.6kJ2{\rm H}_2{\rm O}(g)\rightarrow2{\rm H}_2(g)+{\rm O}_2(g)\;\;\;\;\;\Delta H=483.6\;{\rm{kJ}}
In this reaction, 483.6 kJ of energy must be provided to break down 2 moles of water into 2 moles of hydrogen and 1 mole of oxygen. Note that the ΔH\Delta H is positive, indicating an endothermic reaction. In practical applications of this reaction, energy is provided as electricity, not heat. Enthalpy change also depends on the phase of the reactants and the products. Changing phases requires or releases energy. Changing 1 mole of liquid water into 1 mole of water vapor, for example, absorbs 44 kJ of energy.
H2O(l)→H2O(g)ΔH=44kJ{\rm H}_2{\rm O}(l)\rightarrow{\rm H}_2{\rm O}(g)\;\;\;\;\;\Delta H=44\;{\rm{kJ}}
Step-By-Step Example
Calculation of an Enthalpy Change for the Combustion of Methane
Determine the amount of energy released during the combustion of 12.0 g methane (CH4). The reaction energy is ΔH=−802.3kJ\Delta H=-802.3\;{\rm{kJ}} .
Step 1
Write the balanced equation for the reaction.
CH4(g)+2O2(g)→CO2(g)+2H2O(g)ΔH=−802.3kJ{\rm{CH}}_4(g)+2{\rm{O}}_2(g)\rightarrow{\rm{CO}}_2(g)+2{\rm{H}}_2{\rm{O}}(g)\;\;\;\;\;\;\Delta{H}=-802.3\;{\rm{kJ}}
Step 2
The equation provides enthalpy data for combustion of 1 mole of methane. To calculate how much energy is released by 12.0 grams of methane, convert 12.0 grams of methane gas to moles. First calculate the molar mass of methane. Use the periodic table to determine the atomic mass of each atom. Then, multiply the corresponding molar mass of each atom by the number of atoms shown in the formula for methane.
molar mass of CH4=(1)(12.01g/mol)+(4)(1.01g/mol)=16.05g/mol\begin{aligned}\text{molar mass of }\rm{CH_4}&=(1)(12.01\,{\rm{g/mol}})+(4)(1.01\;{\rm{g/mol}})\\&=16.05\;{\rm{g/mol}}\end{aligned}
Solution
Next, use the molar mass as a conversion factor when determining the energy released by 12.0 g of methane.
ΔH=(ΔHmole)(moles CH4)=(−802.3kJ1molCH4)(12.0gCH4)(1molCH416.0gCH4)=−602kJ\begin{aligned}\Delta H&=(\Delta H_{\rm{mole}})(\text{moles }\rm{CH_4})\\&=\!\left(\displaystyle\frac{-802.3\;{\rm{kJ}}}{1\;{\rm{mol\;CH}}_{4}}\right)\!(12.0\;{\rm{g\;CH}}_{4})\!\left(\displaystyle\frac{1\;{\rm{mol\;CH}}_{4}}{16.0\;{\rm{g\;CH}}_{4}}\right)\\&=-602\;{\rm{kJ}}\end{aligned}
Step-By-Step Example
Determine Enthalpy Change When Forming Ethyne
Ethyne (C2H2), also known as acetylene, forms by adding hydrogen gas (H2) to solid carbon. This reaction is carried out in a constant-pressure calorimeter. When 18.0 grams of carbon reacts with excess hydrogen, 170.0 kJ of heat is absorbed. What is ΔH\Delta H for this reaction?
Step 1
Write the balanced chemical equation for the reaction.
2C(s)+H2(g)→C2H2(g)2{\rm C}(s)+{\rm H}_{2}(g)\rightarrow{\rm C}_{2}{\rm H}_{2}(g)
Step 2
The chemical equation shows what happens when two moles of carbon react with one mole of hydrogen. To solve the problem, first calculate the number of moles of carbon in 18.0 g of carbon.
moles of carbon=mass of carbonmolar mass of carbon=18.0g12.01g/mol=1.499molC\begin{aligned}\text{moles of carbon}&=\frac{\text{mass of carbon}}{\text{molar mass of carbon}}\\&=\frac{18.0\;{\rm{g}}}{12.01\;{\rm{ g/mol}}}\\&=1.499\;{\rm{mol\;C}}\end{aligned}
Solution
Convert the measured heat per 1.499 mol C to ΔH\Delta H for the reaction.
ΔH=(170.0kJ1.499molC)(2molC1molC2H2)=227kJ\begin{aligned}\Delta{H}&=\left(\frac{170.0\;{\rm{kJ}}}{1.499\;{\rm{mol\;C}}}\right)\!\left(\frac{2\;{\rm{mol\;C}}}{1\;{\rm{mol\;C}}_{2}{\rm{H}}_{2}}\right)\\&=227\;{\rm{kJ}}\end{aligned}
The fraction is multiplied by the 2 mol of carbon that is required to produce each mole of C2H2. The final answer is expressed in kJ rather than kJ/mol because ΔH\Delta H is assumed to be per mole.
Scientists measure the enthalpies of reactions experimentally. Consider the combustion of methane:
CH4(g)+2O2(g)→CO2(g)+2H2O(g){\rm{CH}_{4}}(g)+2{\rm{O}_{2}}\;(g)\rightarrow{\rm{CO}_{2}}(g)+2{\rm{H}_{2}}{\rm{O}}(g)
At 1 atm pressure and 25°C, this reaction has a ΔH\Delta H of −802.3 kJ, as given in the question. If the same reaction is carried out at 1,000°C and 1 atm pressure, the ΔH\Delta H changes to −792.4 kJ. Similar differences occur when enthalpy measurements are taken under different conditions. A standardization between enthalpy measurements is needed. The standard state is the set of specific conditions under which reactions are measured, typically 0°C and 1 atm. Recall that 273.15 K is equivalent to 0°C. Standard state is explicitly defined for different substances as follows:
  • For solids and liquids, standard state is a stable state under 1 atm pressure and an often-specified temperature of 25°C.
  • For gases, standard state is a gas with a pressure of 1 atm pressure and an often-specified temperature of 25°C.
  • For solutions, standard state is 1.0 M concentration under 1 atm pressure at an often-specified temperature of 25°C.

When all the reactants and products are at their standard state for a specific reaction, the enthalpy for the reaction is called the standard enthalpy of reaction (ΔH∘\Delta H^\circ).

<Enthalpy>Hess’s Law