Question

Ethanol (c2h5oh) melts at -114c and boils at 78 c the enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vaporization is 38.56 kj/mol. the specific heats of solid and liquid ethanol are 0.97j/g-k and 2.3j/g-k, respectively how much heat is require to convert 75g of ethanol at -120c to the vapor phase at 78c

Answer 1

100 kilo joules There are several phases that this problem undergoes and the final answer is the sum of all the energy used for each phase. Phase 1. Heating of solid ethanol until its melting point. Phase 2. Melting of the ethanol until it's completely liquid. Phase 3. Heating of the liquid ethanol until it reaches its boiling point. Phase 4. Boiling the ethanol until it's completely vapor. To make things more interesting, some of our constant are per gram and some others are per mole. So let's calculate how many moles of ethanol we have. Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass ethanol = 2*12.0107 + 6*1.00794 + 15.999 = 46.06804 g/mol Moles ethanol = 75g / 46.06804 g/mol = 1.628026719 mol Phase 1. Use the specific heat of solid ethanol and multiply by the number of degrees we need to change by the mass we have. So 0.97 J/g*K * 75 g * (-114c - -120c) = 0.97 J/g*K * 75 g * 6K = 436.5 J Phase 2: Time to melt. Just need the moles and the enthalpy of fusion. So: 1.628026719 mol * 5.02 kJ/mol = 8.172694128 kJ Phase 3: Heat to boiling. Just like heating to melting, just a different specific heat and temperature 2.3J/g*K * 75g * (78c - -114c) = 2.3J/g*K * 75g * 192 K = 33120 J Phase 4: Boil it to vapor. Need moles and enthalpy of vaporization. So 1.628026719 mol * 38.56 kJ/mol = 62.77671027 kJ Now let's add them together: 436.5 J + 8.172694128 kJ + 33120 J + 62.77671027 kJ = 0.4365 kJ + 8.172694128 kJ + 33.120 kJ + 62.77671027 kJ =104.5059044 kJ Since the least precise datum we have is 2 significant figures, round the result to 2 significant figures, giving 100 kilo joules.