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 22.8*C. The final temperature of the neutralization reaction was determined to be 41.2°C. The calorimeter constant was known to be 110.53/°C. Density Aqueous Solution - 1.00 g/mL Specific Heat of water = 4.184J/g•°C a. What is the total amount of heat evolved in this reaction? b. If 148 mmol of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization (kl/mole)?

Answer 1

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙^oC), ∆ is a symbol meaning "the change in"

∆T = change in temperature (^oC Celcius)

Energy gained by the acid and base solution:

m = 100 ml = 100 gm C = 4.184 ΔT = 41.2-22.8 = 08.4 °C

q = m C_p ΔT

q = (100 g) (4.184 J/g^-1 °C^-1) (18.4 °C)

q = 7,698.56 J

Energy gained by calorimeter

Q =  18.4 °C x 110.5 J/ °C^-1 = 2,033.2 Joules

Total amount of heat evolved in this reaction =  7,698.56 J + 2033.2 J = 9,731.76‬ J

Molar heat of neutralization = 9,731.76‬ J / 0.148 Mol = 65755.13 Joules / Mole or 65.755 Kilo Joules /mol