6.12
Enthalpies of Reaction
Hess’s law can be used to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are…
For a reaction occurring under standard conditions, a general equation is used to calculate the standard enthalpy change of the reaction. This equation is solved by finding the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants, each multiplied by its stoichiometric coefficient.
Consider the combustion of 2 moles of acetylene gas with 5 moles of oxygen gas to form 4 moles of carbon dioxide gas and 2 moles of water vapor, under standard conditions.
The enthalpy for the reaction equals the sum of four times the enthalpy of formation of carbon dioxide gas and two times the enthalpy of formation of water vapor, minus the sum of two times the enthalpy of formation of acetylene gas and five times the enthalpy of formation of oxygen gas.
The enthalpy equation is derived from combining two concepts: standard enthalpy of formation and Hess’s law.
The first term represents the standard enthalpies of formation of the products; the formation of carbon dioxide from carbon and oxygen — equation 1, and the formation of water from hydrogen and oxygen — equation 2.
The known standard enthalpies of formation for carbon dioxide and water are −393.5 kJ and −241.8 kJ, respectively.
Since the combustion produces 4 moles of carbon dioxide, ΔH1 would be the standard enthalpy of formation of carbon dioxide times 4, which would be −574 kJ.
The combustion also produces 2 moles of water, so the enthalpy change, ΔH2, would be the standard enthalpy of formation of water times 2, which would be −483.6 kJ.
This yields −2058 kJ as the net standard enthalpy of formation of the products.
The second term represents the decomposition of acetylene into carbon and hydrogen — equation 3. This is the reverse of the reaction for the standard enthalpy of formation of the reactant, and thus, its enthalpy value, 227.4 kJ, is preceded by a negative sign, which can also be seen in the enthalpy equation.
Because the reaction consumes 2 moles of acetylene, ΔH3 would be the negative of the standard enthalpy of formation of acetylene times 2, which equals − 53.4 kJ.
Standard formation enthalpy of oxygen is zero.
Therefore, the net standard enthalpy of formation of the reactants is −453.4 kJ.
Recall from Hess’s Law, that if a one-step reaction is carried out in multiple steps, then the sum of the enthalpies of each step, equals the net enthalpy change.
Substituting the values for the enthalpies of formation into the equation gives the enthalpy of the reaction as −2511 kJ.
Q1: How do you calculate the standard enthalpy change of a reaction?
The standard enthalpy change equals the sum of standard enthalpies of formation of products minus the sum of standard enthalpies of formation of reactants, each multiplied by stoichiometric coefficients. This calculation uses standard enthalpy of formation values for all reactants and products under standard conditions to determine the overall enthalpy change for the reaction.
Q2: What is the relationship between Hess's Law and the enthalpy of reaction equation?
Hess's Law states that if a reaction occurs in multiple steps, the sum of enthalpies for each step equals the net enthalpy change. The enthalpy of reaction equation is derived from this principle by decomposing reactants into elements, then recombining them into products. The total enthalpy change is the sum of all intermediate steps combined.
Q3: Why is the standard enthalpy of formation of oxygen zero in reaction calculations?
Oxygen is an element in its standard state, and by definition, the standard enthalpy of formation of an element in its standard state is zero. Since oxygen does not form from simpler elements, it contributes zero to the enthalpy calculation. This applies to all elements in their standard states when calculating reaction enthalpies.
Q4: How do stoichiometric coefficients affect the enthalpy calculation for a reaction?
Stoichiometric coefficients multiply each standard enthalpy of formation value to account for the number of moles of each substance involved. For example, if 4 moles of carbon dioxide form, the enthalpy of formation of CO₂ is multiplied by 4. This ensures the calculated enthalpy reflects the actual quantities of reactants and products in the balanced equation.
Q5: What does a negative enthalpy value indicate about a reaction?
A negative enthalpy value indicates an exothermic reaction that releases heat to the surroundings. For example, the combustion of acetylene yields −2511 kJ, showing significant heat release. Negative values demonstrate that the products have lower enthalpy than the reactants, making the reaction thermodynamically favorable in terms of energy release.
Q6: How do you handle reverse reactions when calculating enthalpy of reaction?
When a reaction is reversed, the sign of its enthalpy value is reversed. For example, decomposing acetylene into carbon and hydrogen is the reverse of acetylene formation, so its enthalpy value becomes negative. This sign convention ensures that the enthalpy equation correctly accounts for whether substances are being formed or decomposed in the overall reaction.
Q7: Can you use the enthalpy of reaction equation for any reaction under standard conditions?
Yes, the enthalpy of reaction equation applies to any reaction occurring under standard conditions if the standard enthalpies of formation for all reactants and products are available. This universal approach allows chemists to predict heat changes for diverse reactions, from combustion to synthesis, without conducting experiments, provided the necessary thermodynamic data exists.