3 Calculations using experimental data

For a reaction in solution we use the following equation energy change = mass of solution x heat capacity x temperature change Q (J) = m (g) x cp (J g-1K-1) x ∆T ( K) This equation will only give the energy for the actual quantities used. Normally this value is converted into the energy change per mole of one of the reactants. (The enthalpy change of reaction, ∆Hr ) Calculating the enthalpy change of reaction, ∆Hr from experimental data General method 1. Using q= m x cp x ∆T calculate energy change for quantities used 2. Work out the moles of the reactants used 3. Divide q by the number of moles of the reactant not in excess to give ∆H 4. Add a sign and unit (divide by a thousand to convert Jmol-1 to kJmol-1 The heat capacity of water is 4.18 J g-1K-1. In any reaction where the reactants are dissolved in water we assume that the heat capacity is the same as pure water. Also assume that the solutions have the density of water, which is 1g cm-3. Eg 25cm3 will weigh 25 g Example 1. Calculate the enthalpy change of reaction for the reaction where 25cm3 of 0.2 M copper sulphate was reacted with 0.01mol (excess of zinc). The temperature increased 7oC . Step 1: Calculate the energy change for the amount of reactants in the test tube. Q = m x cp x ∆T Q = 25 x 4.18 x 7 Q = 731.5 J Step 2 : calculate the number of moles of the reactant not in excess. moles of CuSO4 = conc x vol = 0.2 x 25/1000 = 0.005 mol If you are not told what is in excess, then you need to work out the moles of both reactants and work out using the balanced equation which one is in excess. Step 3 : calculate the enthalpy change per mole which is often called ∆H (the enthalpy change of reaction) ∆H = Q/ no of moles = 731.5/0.005 = 146300 J mol-1 = 146 kJ mol-1 to 3 sf Finally add in the sign to represent the energy change: if temp increases the reaction is exothermic and is given a minus sign e.g. –146 kJ mol-1 Remember in these questions: sign, unit, 3 sig figs. Example 2. 25cm3 of 2M HCl was neutralised by 25cm3 of 2M NaOH. The Temperature increased 13.5oC What was the energy change per mole of HCl? Step 1: Calculate the energy change for the amount of reactants in the test tube. Q = m x cp x ∆T Q = 50 x 4.18 x13.5 Q = 2821.5 J Step 2 : calculate the number of moles of the HCl. moles of HCl = conc x vol = 2 x 25/1000 = 0. 05 mol Step 3 : calculate ∆H the enthalpy change per mole which might be called the enthalpy change of neutralisation ∆H = Q/ no of moles = 2821.5/0.05 = 564300 J mol-1 = -56.4 kJ mol-1 to 3 sf Exothermic and so is given a minus sign Remember in these questions: sign, unit, 3 sig figs. Note the mass equals the mass of acid + the mass of alkali, as they are both solutions. Note the mass is the mass of the copper sulphate solution only. Do not include mass of zinc powder. Example 3. Calculate the enthalpy change of combustion for the reaction where 0.65g of propan-1-ol was completely combusted and used to heat up 150g of water from 20.1 to 45.5oC Step 1: Calculate the energy change used to heat up the water. Q = m x cp x ∆T Q = 150 x 4.18 x 25.4 Q = 15925.8 J Step 2 : calculate the number of moles of alcohol combusted. moles of propan-1-ol = mass/ Mr = 0.65 / 60 = 0.01083 mol Step 3 : calculate the enthalpy change per mole which is called ∆Hc (the enthalpy change of combustion) ∆H = Q/ no of moles = 15925.8/0.01083 = 1470073 J mol-1 = 1470 kJ mol-1 to 3 sf Finally add in the sign to represent the energy change: if temp increases the reaction is exothermic and is given a minus sign eg –1470 kJ mol-1 Remember in these questions: sign, unit, 3 sig figs. Note the mass is the mass of water in the calorimeter and not the alcohol Enthalpies of combustion can be calculated by using calorimetry. Generally the fuel is burnt and the flame is used to heat up water in a metal cup. N Goalby chemrevise.org M