16.) Suppose that following the procedure of Experiment T (Enthalpy and Entropy
ID: 1042216 • Letter: 1
Question
16.) Suppose that following the procedure of Experiment T (Enthalpy and Entropy Changes) you had reacted a 0.950 g sample of MgO with 60.0 mL of 1.2 M HCl according to the reaction AH-7 kJ/mol (1) The initial and final temperatures were 20.0°C and 43.0°C respectively. If you assume th specific heat capacity of the solution was 4.184 /degree-g, and that the density of the solutio 1.00 g/mL, what should you report for ?? (in mol) for reaction (1): (a) 214 kJ/mo (b) -214 kJ/mol (c) 245 kJ/ mol (d) -245 kJ/mol at the -J (e)5.8 kJ/mol
Explanation / Answer
Mass of 60 mL. of HCl solution = Volume * Density = 60 mL. * 1.00 g/ mL. = 60.00 g
Total mass of solution = (60 + 0.950) g = 60.95 g
Temperature rise = (43-20) 0C = 23 0C
Heat associated = mass * specific heat capacity of solution * Temperature rise = 60.95 g * 4.184 J/ g 0C * 23 0C = 5865.34 Joules
Mass of MgO used = 0.950 grams. Moles of MgO used = mass / molar mass of MgO = 0.95 grams/ 40 g/ mole = 0.02375 mole
60 mL of 1.2 M HCl contains 60 * 1.2/1000 mole HCl = 0.072 mole HCl
From the reaction stoichiometry, we see that 0.02375 mole MgO reacts with 2 * 0.02375 mole HCl = 0.0475 mole HCl. So, MgO is the limiting reagent as it completely reacts but ony 0.0475 moles HCl out of 0.072 mole reacts.
0.02375 mole MgO reacts to generate 5865.34 Joules of heat = 5865.34 *10-3 kiloJoules
So , heat per mole = 5865.34 *10-3 kiloJoules/ 0.02375 mole = 245 kJ/ mole
As , the reaction is exothermic (heat is liberated ) so, delH = -245 kJ/ mole
Answer is: (d) -245 kJ/ mole
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