Calculating Enthalpy Change Some reactions occur very slowly! C(s) diamond  C (s) graphite Some reactions occur under conditions that are difficult to duplicate in a laboratory. For these, chemists use a theoretical way to determine the enthalpy change.

Hess’s Law Hess’s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps. 3

A  B + C H = x B + C  D H = y H = ? A  D H = x + y

Figure 6.13: Enthalpy diagram illustrating Hess’s law.

Hess’s Law For example, suppose you are given the following data: Could you use these data to obtain the enthalpy change for the following reaction? 3

Hess’s Law If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third. 3

Example #1 Find H 2CO(g) + 2NO(g)  2CO2(g) + N2(g) a. 2CO(g) + O2(g)  2CO2(g) H=-566.0kJ b. N2(g) + O2(g)  2NO(g) H= 180.6 kJ Reverse b., then cancel c. 2NO(g)  N2(g) + O2(g) H=-566.0 kJ a. 2CO(g) + O2(g)  2CO2(g) H= -180.6 kJ H= -746.6 kJ

Example #2 Find H 4Al(s) + 3MnO2  2Al2O3(s) + 3Mn(s) a. 4Al(s) + 3O2 (g)  2Al2O3(s) H=-3352kJ b. O2(g) + Mn(s) MnO2 (s) H= -521 kJ Reverse b, and multiply by 3. c. 3MnO2 (s)  3O2(g) + 3Mn(s) H=3(521) kJ H= -1789 kJ

Standard Enthalpies of Formation The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atmosphere pressure and the specified temperature (usually 25 oC). The enthalpy change for a reaction in which reactants are in their standard states is denoted DHo (“delta H zero” or “delta H naught”). 4

Standard Enthalpies of Formation The standard enthalpy of formation of a substance, denoted DHfo, is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state. Note that the standard enthalpy of formation for a pure element in its standard state is zero. Note that ONE mole of product can mean fractions of reactants! 4

Standard Enthalpies of Formation The law of summation of heats of formation states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants. S is the mathematical symbol meaning “the sum of”, and m and n are the coefficients of the substances in the chemical equation. 4

Example #1 Large quantities of ammonia are used to prepare nitric acid according to the following equation: What is the standard enthalpy change for this reaction? Use Table 16.7 for data. 4

Example #1 Large quantities of ammonia are used to prepare nitric acid according to the following equation: What is the standard enthalpy change for this reaction? Use Table 16.7 for data. 4

Example #1 You record the values of DHfo under the formulas in the equation, multiplying them by the coefficients in the equation. You can calculate DHo by subtracting the values for the reactants from the values for the products. 4

Example #1 Using the summation law: Be careful of arithmetic signs as they are a likely source of mistakes. 4

Example #2 You record the values of DHfo under the formulas in the equation, multiplying them by the coefficients in the equation. 2(-273) +(-1220) – [(-21) + 4(0)] Horxn = -1745 kJ **The value for an element (even diatomic) is zero. Remember to subtract the reactants FROM the products.**Remember to multiple the Hf values by the coefficients. 4

Example #3 Calculate the enthalpy change from standard enthalpies of formation. CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) Products: CO2(g) = -394 kJ and 2H2O(l)= 2(-286) Reactants: CH4(g) = -75 kJ and 2O2(g) = 2 (0) [-394 + 2(-286)]- [-75] = 891 kJ