6. 1. 1 If the reaction produces heat (increases the temperature of the surroundings) then it’s exothermic. If it decreases the temp (i. e. absorbs heat) then it’s endothermic. Also, the yield of an equilibrium reaction which is exothermic will be increased if it occurs at low temps, and so for endothermic reactions at high temperatures. 6. 1. 2 Exothermic : A reaction which produces heat. Endothermic : A reaction which absorbs heat. Enthalpy of reaction : The change in internal energy (H) through a reaction is ? H. 6. 1. 3 H will be negative for exothermic reactions (because internal heat is being lost) and positive for endothermic reactions (because internal energy is being gained). 6. 1. 4 The most stable state is where all energy has been released. Therefore when going to a more stable state, energy will be released, and when going to a less stable state, energy will be gained. On an enthalpy level diagram, higher positions will be less stable (with more internal energy) therefore, if the product is lower, heat is released (more stable, ? H is negative) but if it is higher, heat is gained (less stable, ?
H is positive). 6. 1. 5 Formation of bonds : Release of energy. Breaking of bonds : Gain / absorption of energy. 6. 2 Calculation of enthalpy changes 6. 2. 1 Change in energy = mass x specific heat capacity x change in temperature ? (E = m x C x ? T) 6. 2. 2 Enthalpy changes (? H) are related to the number of mols in the reaction. If all the coefficients are doubled, then the value of ? H will be doubled. Attention must be paid to limiting reagents though, because enthalpy changes depend on the amount of reactants reacted (extensive property of enthalpy). . 2. 3 When a reaction is carried out in water, the water will gain or lose heat from (or to) the reaction, usually with little escaping the water. Therefore, the change in energy, and so the ? H value, can be calculated with E = m x c x ? T where E is equal to ? H, m is the mass of water present, and c = 4. 18 kJ Kg-1 K-1. This ? H value can then be calculated back to find the enthalpy change for each mol of reactants. 6. 2. 4 The solution should be placed in a container as insulated as possible, to keep as much heat as possible from escaping.
The temperature should be measured continuously , and the value used in the equation is the maximum change in temp from the initial position. 6. 2. 5 The results will be a change in temperature. This can be converted into a change in heat (or energy) by using the above equation and a known mass of water. This can be used to calculate the ? H for the amount of reactants present, which can then be used to calculate for a given number of mols. 6. 3 Hess’ Law 6. 3. 1 Hess’ Law states that the total enthalpy change between given reactants and products is the same regardless of any intermediate steps (or the reaction pathway).
To calculate: ?Reverse any reactions which are going the wrong way and invert the sign of their ? H values. ?Divide or multiply the reactions until the intermediate products will cancel out when the reactions are vertically added (always multiply/divide the ? H value by the same number). ?Vertically add them. ?Divide or multiply the resulting reaction to the correct coefficients. 6. 4 Bond enthalpies 6. 4. 1 Bond enthalpy (aka dissociation enthalpy) : The enthalpy change when one mol of bonds are broken homolitically in the gas phase. i. e. X-Y(g) -> X(g) + Y(g) : ? H(dissociation).
Molecules such as CH4 have multiple C-H bonds to be broken, and so the bond enthalpy for C-H is actually an average value. These values can be used to calculate unknown enthalpy changes in reactions where only a few bonds are being formed/broken. 6. 4. 2 If the reaction can be expressed in terms of the breaking and formation of bonds in a gaseous state, then by adding (or subtracting when bonds are formed) the ? H values the total enthalpy of reaction can be found. 16. 1 Standard enthalpy changes of reaction 16. 1. 1 Standard state : 101 kPa, 298 K (or 1 atm, 25 degrees celcuis).
Standard enthalpy change of formation : The enthalpy change when 1 mol of a substance is made from its elements in their standard states. For example C(graphite) + 2H2(g) -> CH4(g). Molecules, like H2 are considered to be ‘standard state’. Fractions of mols (i. e. fractions in coefficients), may also be used if necessary as 1 mol must be produced). 16. 1. 2 If a reaction can be expressed in terms of changes of formation (and bond enthalpies as in SL) then add up all the ? H values to get the ? H for the reaction. 16. 2 Lattice enthalpy 16. 2. 1
Lattice enthalpy : The enthalpy change when 1 mol of crystals (i. e. an ionic lattice) is formed from its component particles at an infinite distance apart. M+(g) + X-(g) -> MX(s) The value of lattice enthalpy is assumed to be positive for the separation of the lattice, and negative for the formation of the lattice. 16. 2. 2 As above, lattice enthalpies just add another type of reaction to those which can be shown on the Born-Haber cycle. 16. 2. 3 Lattice enthalpy increases with higher ionic charge and with smaller ionic radius (due to increased attraction). 6. 3 Entropy 16. 3. 1 Factors which increase disorder in a system: ?Mixing of particles. ?Change of state to greater distance between particles (solid -> liquid or liquid -> gas). ?Increased particle movement (temperature). ?Increased number of particles (when more gas particles are produced, this generally outweighs all other factors). 16. 3. 2 Predict the sign of ? S (the change in entropy) for a reaction based on the above factors. ?S is positive when entropy increases (more disorder) and negative when entropy decreases (less disorder). 16. 3. 3
The standard entropy change can be calculated by subtracting the absolute entropy of the reactants from that of the products. 16. 4 Spontaneity of a reaction 16. 4. 1 Reactions which release heat (and so increase stability) tend to occur as do reactions which increase entropy (? S is positive). Neither of these can be used to accurately predict spontaneity alone however. 16. 4. 2 When ? G is negative, the reaction is spontaneous, when it’s positive, the reaction is not. 16. 4. 3 ?G = ? H – Temperature(in kelvin) x ? S Spontaneity depends on ? H, ? S and the temperature at which the reaction takes place (or doesn’t as the case may be). 6. 4. 4 Substitute values into the equation above. Hopefully that’s not too tricky. 16. 4. 5 There are four possibilities: 1.? G is always negative when ? H is negative and ? S is positive. 2.? G is negative at high temperatures if ? H is positive and ? S is positive (i. e. an endothermic reaction is spontaneous when T x ? S is greater than ? H). 3.? G is negative at lower temperatures if ? H is negative and ? S is negative (exothermic reactions are spontaneous if ? H is bigger than T x ? S). ?G is never negative if ? H is positive and ? S negative.