Question

4. You mix 50.0 ml. of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions and the calorimeter were initially at 23.9°C. The final temperature of the neutralization reaction was determined to be 41.2°C. The calorimeter constant was known to be 110.6J/°C. Density Aqueous Solution = 1.00 g/ml Specific Heat of water = 4.184 J/g °C Hide a. What is the total amount of heat evolved in this reaction? NG 3 b. If 149 mmol of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization (kJ/mole)? NG

Answer 1

mL MASS of Solton = (50.0 + 50.0) ml x 1.00G = 100.00 a) HEAT OF REACTION - HEAT ABSORBED BY WATER + HEAT ABSORBED BY CALORIMETER MCAT + Case AT = 100.00x 4.184 I (41.2 – 23.9) < + 110.6 Į * (41.2 - 23.8) 110 = 19.15 kJ 6 molar CAR HEAT OF HEAT OF NEUTRAL ZATION = A - 9.15 KT 149 mmolx 0.001 mot 1 mmol RJ = |-61.41 here mol MINUS SIGN INDY CARES REACTION IS EXOTHERMIC