Standard enthalpy changes and Hess’ Law

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Standard enthalpy changes and Hess’ Law d. recall the definition of standard enthalpy changes of reaction, formation, combustion, neutralization and atomization and use experimental data to calculate energy transferred in a reaction and hence the enthalpy change of the reaction. This will be limited to experiments where substances are mixed in an insulated container, and combustion experiments e. recall Hess’s 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 g. demonstrate an understanding of the terms bond enthalpy and mean bond enthalpy, and use bond enthalpies in Hess cycle calculations and recognise their limitations. Understand that bond enthalpy data gives some indication about which bond will break first in a reaction, how easy or difficult it is and therefore how rapidly a reaction will take place at room temperature. Home Learning Task – Read pp43-45 and answer the questions therein. Crowe2010

100 cm3 of water (100g) was measured into the calorimeter. Connector - Calculations using the results from an experimental calorimeter. 100 cm3 of water (100g) was measured into the calorimeter. The spirit burner contained the fuel ethanol C2H5OH and weighed 18.62g at the start. After burning it weighed 17.14g and the temperature of the water rose from 18 to 89oC. Method Use q = mcΔT to calculate the energy released. ( c = 4.2 J g-1 K-1 ) q represents the heat released from the mass of ethanol burnt. Calculate the energy released per g. Calculate the energy released per mole Now give the Enthalpy of Combustion of ethanol (ΔHc) in kJ mol-1.

The temperature rise = 89 - 18 = 71oC (exothermic). Mass of fuel burned = 18.62-17.14 = 1.48g. Heat absorbed by the water = mass of water x SHCwater x temperature = 100 x 4.2 x 71 = 29820 J (for 1.48g) heat energy released per g = energy supplied in J / mass of fuel burned in g heat energy released on burning = 29820 / 1.48 = 20149 J/g of C2H5OH this energy change can be also expressed on a molar basis. Relative atomic masses Ar: C = 12, H = 1, O = 16 Mr(C2H5OH) = (2 x 12) + (1 x 5) + 16 + 16 = 46, so 1 mole = 46g Heat released (given out) by 1 mole of C2H5OH = 46 x 20149 = 926854 J/mole or 927 kJ/mol (3 sf) Enthalpy of combustion of ethanol = ΔHc = -927 kJmol-1

Standard enthalpy change of ….. Formation ΔHθf is the enthalpy change when 1 mole of a compound is made from its elements in their standard states, under standard conditions. Combustion ΔHθc is the enthalpy change when 1 mole of a substance burns completely in oxygen, under standard conditions. Atomisation ΔHθat is the enthalpy change when 1 mole of gaseous atoms is made from the element in its standard state, under standard conditions. Reaction ΔHθr is the enthalpy change when molar quantities of reactants as stated in an equation react under standard conditions. Neutralisation ΔHθn is the energy released when unit molar quantities of acids and alkalis completely neutralise each other at 298K. Note: Make sure that you know these definitions

Hess’s Law The enthalpy change in turning any reactants into a set of products is the same no matter what route we take. C c b a A B d e D a = b + c = d - e Note: arrows in same direction as “a” are added, Those in the opposite direction are subtracted.

Using enthalpy cycles Some enthalpy changes cannot be measured directly: e.g. The enthalpy of formation of ethanol Write the equation for the formation of ethanol from its elements. Why can’t ethanol be prepared in this way? Standard Enthalpy of Formation, ΔHθf is the enthalpy change when 1 mole of a compound is made from its elements in their standard states, under standard conditions.

Using enthalpy cycles to calculate enthalpy changes that cannot be measured directly Although we cannot directly measure the enthalpy of formation of ethanol, it can be calculated by using other measurable enthalpy changes. Where: ΔH1 = 2x (enthalpy of combustion of graphite) + 3x (enthalpy of combustion of hydrogen) ΔH3 = enthalpy of combustion of ethanol These enthalpies of combustion can be measured, and so ΔH1 can be calculated since, applying Hess’ Law: ΔH1 = ΔH2 – ΔH3

Enthalpy of Formation of Ethanol Use the information above to calculate the enthalpy of formation of ethanol.

Calculating the heat of formation of ethanol From the previous slide: ΔH1 = ΔH2 – ΔH3 2C + 3H2 + ½ O2 C2H5OH So: ΔH2 = 2ΔHcθ[C, graphite(s)] + 3ΔHcθ[H2,(g)] ΔH3 = ΔHcθ[C2H5OH(l)] ΔH1 = ΔH2 – ΔH3 = [2(-393.5) + 3(-285.8)] – [-1367.3] = [-787 – 857.4] + 1367.3 = -1644.4 + 1367.3 = -277.4kJmol-1

Enthalpy Level Diagram This is another way to show the enthalpy changes in an energy cycle:

Ca(s) + C(graphite) + 1½O2(g) ➔ CaCO3(s) Practical 1.6 Finding an enthalpy change that cannot be measured directly The standard enthalpy of formation of calcium carbonate, ΔHf [CaCO3] is defined as the enthalpy change when one mole of calcium carbonate is formed from its elements in their standard states, under standard conditions. It is given by the following equation: Ca(s) + C(graphite) + 1½O2(g) ➔ CaCO3(s) The reaction, as written, is unlikely to take place in the laboratory, and so ΔHf [CaCO3] cannot be determined directly. However, the enthalpy changes when calcium metal and calcium carbonate react with hydrochloric acid, are measurable, and applying the principles of Hess’s law, ΔHf [CaCO3] can then be determined.

Practical 1.6 – Finding an enthalpy change that cannot be measured directly

Practical 1.6 Finding an enthalpy change that cannot be measured directly

Useful information Usually one reagent is in excess to ensure a complete reaction So calculations should be based on the fully reacted reagent. Certain assumptions are made during the calculation The density of the solution and its specific heat capacity is that of water. That no heat is lost to the surroundings. The main source of error in these experiments is heat loss to the surroundings – (atmosphere & equipment) Other sources of error include: incorrect measurements solution concentrations mass of reactants 15

Worksheets Exp 1.6 from teacher’s guide