You mix 50.0 mL of a weak monoprotic acid wilh 50.0 ml. of NaOH solution in a co

Question

You mix 50.0 mL of a weak monoprotic acid wilh 50.0 ml. of NaOH solution in a coffee cup calorimeter. Both solutions and the calorimeter were initially a temperature 22.0degreesC.The final temperature of the neutralization reaction was determined to be41.2degreeC.the calorimeter constant was known to be 109.2 j/degreeC. (Specific heat of H_2.O = 4.184J |/g degreeC). Show all work. 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?

Explanation / Answer

Solution :-

Total volume of solution = 50 ml + 50 ml = 100 ml

Mass of solution = 100 g assuming density = 1 g / ml

Initial temperature = 22.0 C

Final temperature = 41.2 C

S cal = 109.2 J/C

Specific heat of water = 4.184 J per g C

a) Total amount of heat given by reaction

delta T = 41.2 C – 22.0 C = 19.2 C

Q total = (S cal * delta T) +(m*S*delta T)

              = (109.2 J per C * 19.2 C) + (100 g * 4.184 J per g C * 19.2 C)

             = 10130 J

So the total heat given by reaction (evolved) = 10130 J

b)Molar heat of neutralization

155 mmol * 1 mol / 1000 mmol = 0.155 mol

Molar heat of neutralization = q total / moles

                                                    = 10130 J / 0.155 mol

                                                   = 65355 J/mol

65355 J per mol * 1 kJ / 1000 J = 65.355 kJ/mol

Since reaction is exothermic so the

Molar heat of neutralization have negative sign

So the molar heat of neutralization = - 65.355 kJ /mol

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