Enthalpy Change(ΔH)

Enthalpy ChangeKey Concepts The heat content of a chemical reaction is called the enthalpy (symbol: H)

The enthalpy change (ΔH) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure. ΔH = H(products) - H(reactants) The units are usually given as kJ mol-1 (kJ/mol) or sometimes as kcal mol-1 (kcal/mol)1 calorie (1 cal) = 4.184 joules (4.184 J) Standard enthalpy change (ΔHѲ);Energy changes for any reaction which are measured under standard conditions of temperature and pressure25oC (298K) - 101.3 kPa (1 atmosphere) - 1 mol dm-3

Manipulating the Enthalpy Change Term

N2(g) + 3H2(g) -----> 2NH3(g)       ΔH = - 92.4 kJ/mol

92.4 kJ of energy is released for every 1 mole of N2(g) 92.4 kJ of energy is released for every 3 moles of H2(g) 92.4 kJ of energy is released for every 2 moles of NH3(g) produced.

How much energy is released if only 1 mole of ammonia (NH3) gas is produced?

a) Standard Enthalpy of formation (∆Hf0) It is the enthalpy change when

one mole of a compound is formed from its elements in their standard states under standard conditions of temperature and pressureEg: C(s) + O2 (g) → CO2 (g) ∆Hf

0 =-393.5 KJ

practice question N2(g) + 3H2(g) -----> 2NH3(g)       ΔH0 = - 92.4 kJ/molFind enthalpy of formation ∆Hf

0 of NH3 from the equation (remember enthalpy of formation ∆Hf

0 is always for 1 mole)

∆Hf0 of a free element in its standard state is taken as zero.

Eg: ∆Hf0 = 0 for H2(g) , N2 (g), O2(g) , F2(g), Cl2(g), Br2(l), I2 (s) ,

C(s,graphite)

How to find ∆H0 from std. enthalpy of formation

∆H0 = Σ ∆Hf0 (products) − Σ ∆Hf

0 (reactants)

Calculate enthalpy change for the reaction

CH4 +2O2(g) →CO2(g) +2H2O(g)

∆Hf0 of CH4 , CO2(g) , H2O(g, are -74.8, -393.5 and -285.8 kjmol-1 respectively

Enthalpy of combustion (∆Hc0) : It is enthalpy change when one

mole of a substance is completely burnt in oxygen under standard conditions of temperature and pressure

Eg: CH4 (g) +2O2 (g) →CO2 (g) + 2H2O (g) ∆H = – 890.3 KJ

Enthalpy of solution (∆H0soln): The enthalpy change when one mole

of substance is dissolved in water.

Eg: NaOH (s) + H2O -----> NaOH(aq) ∆Hsoln = -441.1KJ/mol

Enthalpy of neutralization : The enthalpy change when one mole of acid (H+) is neutralized with one mole of base(OH-) to form one mole of

water Eg: HCl (aq) + NaOH (aq) → NaCl(aq) + H2O (l) + Heat

H+ (aq) + OH- (aq) → H2O (l) ∆H0 = –57.1 KJ/mol The enthalpy of neutralization of all strong acids by strong bases is

always equal to -57.1 KJ .

Practice Problem:

Enthalpy of neutralization• To find the enthalpy of neutralization for acid base reactions from

experimental data. we use this rule.

Heat (J) = mass of solution(g) x Specific heat capacity (JK-1g-1) x change in temperature (K or C)

Q = m x c x ∆ TQuestions1. 50.0 cm3 of 1.00 mol dm-3 HCl solution was added to 50.0 cm3 of 1.00 mol

dm-3 sodium hydroxide solution in a polystyrene beaker. The initial temperature of both solutions was 16.7 °C. after stirring the highest temperature reach was 23.5 °C. calculate the enthalpy change for this reaction. ( Specific heat capacity of water is 4.2JK-1g-1)

Answer: -57.1KJ mol-1

2. Chemistry book page 309 questions 1- 2

Practice problem A student used simple calorimeter to determine the enthalpy change for the

combustion of ethanolC2H5OH + 3O2 → 2CO2 +3H2O

when 0.690g of ethanol was burnt, it produced a temperature rise

of 13.2K in 250g of water. Calculate ∆H for the reaction

( Specific heat capacity of water is 4.2JK-1g-1)