Enthalpy change

1. Enthalpy Change ( Δ H)

2. Enthalpy Change Key Concepts The heat content of a chemical reaction is called the enthalpy (symbol: H) The enthalpy change ( Δ H) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure. Δ H = 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 ( Δ H Ѳ ) ; Energy changes for any reaction which are measured under standard conditions of temperature and pressure 25 o C (298K) - 101.3 kPa (1 atmosphere) - 1 mol dm -3

3. Manipulating the Enthalpy Change Term N 2 (g) + 3H 2 (g) -----> 2NH 3 (g)       Δ H = - 92.4 kJ/mol 92.4 kJ of energy is released for every 1 mole of N 2 (g) 92.4 kJ of energy is released for every 3 moles of H 2 (g) 92.4 kJ of energy is released for every 2 moles of NH 3 (g) produced. How much energy is released if only 1 mole of ammonia (NH 3 ) gas is produced?

4. a) Standard Enthalpy of formation ( ∆H f ) 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 pressure Eg : C(s) + O 2 (g) → CO 2 (g) ∆H f =-393.5 KJ practice question N 2 (g) + 3H 2 (g) -----> 2NH 3 (g)       Δ H = - 92.4 kJ/mol Find enthalpy of formation ∆H f of NH3 from the equation (remember enthalpy of formation ∆H f is always for 1 mole) ∆H f of a free element in its standard state is taken as zero. Eg : ∆H f = 0 for H 2 (g) , N 2 (g), O 2 (g) , F 2 (g), Cl 2 (g), Br 2 ( l ), I 2 (s) , C( s,graphite ) How to find ∆H from std. enthalpy of formation ∆H = Σ ∆H f (products) − Σ ∆H f (reactants)

5. Calculate enthalpy change for the reaction CH 4 +2O 2 (g) →CO 2 (g) +2H 2 O(g) ∆H f of CH 4 , CO 2 (g) , H 2 O(g, are -74.8, -393.5 and -285.8 kjmol-1 respectively

6. Enthalpy of combustion ( ∆H c ) : It is enthalpy change when one mole of a substance is completely burnt in oxygen under standard conditions of temperature and pressure Eg : CH 4 (g) +2O 2 (g) →CO 2 (g) + 2H 2 O (g) ∆H = – 890.3 KJ Enthalpy of solution ( ∆H soln ) : The enthalpy change when one mole of substance is dissolved in water. Eg : NaOH (s) + H2O -----> NaOH ( aq ) ∆ H soln = - 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 ) + H 2 O ( l ) + Heat H+ ( aq ) + OH- ( aq ) → H 2 O ( l ) ∆H = –57.1 KJ/mol The enthalpy of neutralization of all strong acids by strong bases is always equal to -57.1 KJ . Practice Problem:

7. 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 -1 g -1 ) x change in temperature (K or C) Q = m x c x ∆ T Questions 50.0 cm 3 of 1.00 mol dm -3 HCl solution was added to 50.0 cm 3 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 -1 g -1 ) Answer: - 57.1KJ mol -1 2. Chemistry book page 309 questions 1- 2

8. Practice problem A student used simple calorimeter to determine the enthalpy change for the combustion of ethanol C2H5OH + 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 -1 g -1 )