5.1 Exothermic and endothermic reactions5.1 Exothermic and endothermic reactions

If a reaction produces heat (increases the If a reaction produces heat (increases the temperature of the surroundings) it is exothermic.temperature of the surroundings) it is exothermic. If the temperature of the reaction mixture If the temperature of the reaction mixture decreases (ie heat is absorbed) then the reaction is decreases (ie heat is absorbed) then the reaction is endothermic. endothermic.

Exothermic ->Exothermic -> a reaction which produces heat a reaction which produces heat (ΔH has a negative value by convention, -ve)(ΔH has a negative value by convention, -ve)

Endothermic ->Endothermic -> a reaction which absorbs heat a reaction which absorbs heat (ΔH has a positive value by convention, +ve)(ΔH has a positive value by convention, +ve)

Enthalpy of reaction, Enthalpy of reaction, ΔHΔH:: The change in internal (chemical) energy The change in internal (chemical) energy

(H) in a reaction(H) in a reaction

The most stable state is where all energy has The most stable state is where all energy has been released. When going to a more stable been released. When going to a more stable state, energy will be released, and when going to state, energy will be released, and when going to a less stable state, energy will be gained (from a less stable state, energy will be gained (from the surroundings). the surroundings).

On an enthalpy level diagram, higher positions On an enthalpy level diagram, higher positions will be less stable (with more internal energy) will be less stable (with more internal energy) therefore, if the product is lower, heat is released therefore, if the product is lower, heat is released (more stable, ΔH is -ve) but if it is higher, heat is (more stable, ΔH is -ve) but if it is higher, heat is gained (less stable, ΔH is +ve).gained (less stable, ΔH is +ve).

Formation of bonds causes an Formation of bonds causes an energy release (exothermic). energy release (exothermic).

Breaking of bonds requires Breaking of bonds requires energy (endothermic).energy (endothermic).

Calculation of enthalpy changesCalculation of enthalpy changes

Change in energy = mass x specific Change in energy = mass x specific heat capacity x change in heat capacity x change in temperature temperature

E = m x c x ΔTE = m x c x ΔT

where m is the mass of water present where m is the mass of water present (kilograms), and c = 4.18 kJ Kg-1 K-1. (kilograms), and c = 4.18 kJ Kg-1 K-1.

Hess' Law Hess' Law

Hess' Law states that the total Hess' Law states that the total enthalpy change between given enthalpy change between given reactants and products is that same reactants and products is that same regardless of any intermediate steps regardless of any intermediate steps (or the reaction pathway). (or the reaction pathway).

Example: Calculate the enthalpy of Example: Calculate the enthalpy of formation of methane. formation of methane.

Example 1:Example 1:

Data:Data:

ΔH combustion (carbon) = -394kj ΔH combustion (carbon) = -394kj

ΔH combustion (hydrogen) =-242kj ΔH combustion (hydrogen) =-242kj

ΔHcombustion (methane) =-891kj ΔHcombustion (methane) =-891kj

If If ΔHf methane is represented by the equation:ΔHf methane is represented by the equation: C(s) + H2(g) C(s) + H2(g) →→ CH4 (g) CH4 (g) This equation can be constructed using the equations for combustion of This equation can be constructed using the equations for combustion of

the reactants (carbon and hydrogen) and the product (methane)the reactants (carbon and hydrogen) and the product (methane) C(s) + O2(g) C(s) + O2(g) → → CO2(g) -394k CO2(g) -394k……..enthalpy 1..enthalpy 1     H2(g) + H2(g) + ½½ O2(g) O2(g) →→ H2O(l) -242kj H2O(l) -242kj……..enthalpy 2..enthalpy 2multiply this equation by 2 to getmultiply this equation by 2 to get      2H2(g) + O2(g) 2H2(g) + O2(g) →→ 2H2O(l) -484kj 2H2O(l) -484kj ………….enthalpy 3.enthalpy 3

Add the first and third equation together to get:Add the first and third equation together to get:  C(s) +2O2(g) +2H2(g) C(s) +2O2(g) +2H2(g) →→CO2(g) + 2H2O(l) -878kj CO2(g) + 2H2O(l) -878kj enthalpy (1 + 3) = 4enthalpy (1 + 3) = 4

Now take away the eq for the combustion of methaneNow take away the eq for the combustion of methane  CH4 (g) + 2O2(g) CH4 (g) + 2O2(g) →→ CO2(g) + 2H2O(l) -891kj enthalpy 5 CO2(g) + 2H2O(l) -891kj enthalpy 5    

And after rearrangement ( take the CH4 to the right hand side) the And after rearrangement ( take the CH4 to the right hand side) the result is the equation for the formation of methaneresult is the equation for the formation of methane  

C(s) + H2(g) C(s) + H2(g) →→ CH4 (g) +13kj CH4 (g) +13kj enthalpy ( 4 - 5)enthalpy ( 4 - 5)

Example 1Example 1Monoclinic sulphur is formed in volcanic regions by reaction between sulphur Monoclinic sulphur is formed in volcanic regions by reaction between sulphur

dioxide and hydrogen sulphide according to the equation:dioxide and hydrogen sulphide according to the equation: SO2 + 2H2S --> 2H2O + 3SSO2 + 2H2S --> 2H2O + 3SDraw an enthalpy diagram or cycle and calculate the standard enthalpy change for this Draw an enthalpy diagram or cycle and calculate the standard enthalpy change for this

reaction.reaction. Here are some values.Here are some values. Standard enthalpy of formation (all in kJ)Standard enthalpy of formation (all in kJ) HH22O(l) -286 O(l) -286 HH22S(g) - 20.2 S(g) - 20.2 Standard enthalpy of combustion of S (monoclinic) = -297.2kJStandard enthalpy of combustion of S (monoclinic) = -297.2kJ

equation 1: Hequation 1: H22 + + ½½ O O22 --> H --> H22OO         -286 kJ -286 kJ equation 2: Hequation 2: H22 + S --> H2S + S --> H2S          -20.2kJ -20.2kJ equation 3: S + Oequation 3: S + O22 --> SO2 --> SO2          -297.2 kJ -297.2 kJ multiply E2 by 2multiply E2 by 2…… (equation 4) H (equation 4) H2 2 + 2S --> 2H+ 2S --> 2H22S S        -40.4 kJ -40.4 kJ add equation 3 & 4add equation 3 & 4……..…….. 2H.. 2H22 + 3S + O + 3S + O22 --> SO --> SO22 + 2H + 2H22SS          -337.6 kJ -337.6 kJ multiply E1 by 3multiply E1 by 3……(equation 5) 2H(equation 5) 2H22 + O + O2 2 --> 2H--> 2H22OO       -572 kJ -572 kJ subtract 4 from 5subtract 4 from 5…………-3S ---> 2H2O - SO2 - 2H2S -3S ---> 2H2O - SO2 - 2H2S          -234.4 kJ -234.4 kJ

rearrangerearrangeSOSO22 + 2H + 2H22S --> 2HS --> 2H22O + 3SO + 3S         -234.4 kJ-234.4 kJ

Born Haber Cycles Born Haber Cycles Born Haber cycles are the application of Born Haber cycles are the application of

Hess' law to ionic systems. An ionic solid Hess' law to ionic systems. An ionic solid consists of a giant structure of ions held consists of a giant structure of ions held together in a giant lattice.together in a giant lattice.

Application of Hess law tells us that the Application of Hess law tells us that the enthalpy of formation of an ionic crystal is enthalpy of formation of an ionic crystal is equal to the sum of the energies of equal to the sum of the energies of formation of the ions plus the enthalpy of formation of the ions plus the enthalpy of the lattice. the lattice.

it is a several step process that is best it is a several step process that is best represented by a diagram showing the represented by a diagram showing the individual steps as individual steps as endothermic endothermic upwards upwards andand exothermic downwards. exothermic downwards.