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</li></li></ul><li>Manipulating the Enthalpy Change Term<br />N2(g) + 3H2(g) -----> 2NH3(g) ΔH = - 92.4 kJ/mol <br /><ul><li> 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. </li></ul>How much energy is released if only 1 mole of ammonia (NH3) gas is produced? <br />
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) ∆Hf0 =-393.5 KJ practice question N2(g) + 3H2(g) -----> 2NH3(g) ΔH0 = - 92.4 kJ/molFind enthalpy of formation ∆Hf0of NH3 from the equation (remember enthalpy of formation ∆Hf0 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 ∆Hf0 (reactants)<br />
Calculate enthalpy change for the reactionCH4 +2O2(g) ->CO2(g) +2H2O(g) ∆Hf0of CH4 , CO2(g) , H2O(g, are -74.8, -393.5 and -285.8 kjmol-1 respectively<br />
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/molThe enthalpy of neutralization of all strong acids by strong bases is always equal to -57.1 KJ .Practice Problem:<br />
Enthalpy of neutralization<br />To find the enthalpy of neutralization for acid base reactions from experimental data. we use this rule.<br />Heat (J) = mass of solution(g) x Specific heat capacity (JK-1g-1) x change in temperature (K or C)<br /> Q = m x c x ∆T<br />Questions<br />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) <br />Answer: -57.1KJ mol-1<br />2. Chemistry book page 309 questions 1- 2<br />
Practice problem A student used simple calorimeter to determine the enthalpy change for the combustion of ethanolC2H5OH + 3O2 -> 2CO2 +3H2Owhen 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) <br />
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