4. You mix 50.0 mL of a weak monoprotic acid with 50.0 ml. of NaOH solution in a

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 a temperature of 22.0 °C. The fin temperature of the neutralization reaction was determined to be 41.2 °C. The calorimeter constant was known to be 109.2 Jr. (specific heat of H,O = 4.1841/ goC). Show all work. al a. What is the total amount of heat evolved in this reaction? If 155 mmol of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization for this reaction? b.

Explanation / Answer

4. For the given neutralization reaction

(a) heat = mCpdT

total heat evolved = 100 x 4.184 x 19.2 + 109.2 x 19.2 = 10.130 kJ

(b) 155 mmol of monoprotic acid neutralized

molar heat of neutralization = -10.130/0.155 = 65.355 kJ/mol

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