# 1.4 Energetics Practical 1.6 - Finding an enthalpy change ... 1.4 Energetics Practical 1.6 Finding an enthalpy change that cannot be measured directly e. recall Hesss Law and apply it to calculating enthalpy changes of reaction from data provided, selected from a table of data or obtained from experiments and understand why standard data is necessary to carry out calculations of this type Connector: 1. State Hesss Law 2. Define the standard enthalpy of formation 3. What are standard conditions? 4. What symbol is used to indicate standard conditions? Crowe2009 Hesss Law

Hesss Law states that the enthalpy change in turning any reactants into a set of products is the same no matter what route we take. b C c a A d D

B e a=b+c=d-e Note: arrows in same direction as a are added, Those in the opposite direction are subtracted. indicates standard conditions Enthalpy Cycle Calculation The enthalpy of formation of ethanol cannot be measured directly. However, the standard enthalpy of combustion for graphite, hydrogen and ethanol are known.

Construct an energy cycle and then use the data below to calculate the enthalpy of formation of ethanol. Hint: Start by writing the equation for the formation of ethanol from its elements 2C (graphite, s) + 3H2 (g) + O2 (g) H1 C2H5OH (l) (+ 3O2 (g)) (+ 3O2 (g)) H3

H2 2CO2 (g) + 3H2O (l) Note the 3O2 (g) s cancel each other out and so can be omitted from the cycle H1 = H2 H3 Calculating the heat of formation of ethanol From the previous slide: H1 = H2 H3 2C + 3H2 + O2 C2H5OH

So: H2 = 2Hc[C, graphiteC, graphite(s)] + 3Hc[C, graphiteH2,(g)] H3 = Hc[C, graphiteC2H5OH(l)] H1 = H2 = [C, graphite2(-393.5) + 3(-285.8)] = [C, graphite-787 857.4] + 1367.3 = -1644.4 + 1367.3 = -277.4kJmol-1 H3 [C, graphite-1367.3]