2. A styrofoam cup has a calorimeter constant of 9.8 cal / C. Show your work for

Question

2.

A styrofoam cup has a calorimeter constant of 9.8 cal / C. Show your work for each answer.
(a) If a neutralization reaction performed in this calorimeter causes the temperature to rise from 22.3 C to 36.5 C, how many calories of heat were absorbed by the cup?
(b) In the same calorimeter, a neutralization reaction was carried out wherein the m ass of the solution was 100. g with a specific heat of 1.03 cal / gm C. If the temperature rose from 22.3 C to 36.5 C, calculate the number of calories of heat energy absorbed by the solution.

(c) What was the heat of neutralization of the reaction?
(d) If 50.0 ml of 2.0 M acid was used in the reaction, how many moles of acid were present? (e) What is the molar heat of neutralization for this reaction? (answer in kcal / mol)

Explanation / Answer

(a) Rise in temperature of the calorimeter cup = (36.5 – 22.3)°C = 14.2°C.

The cup has a calorimeter constant 9.8 cal/°C. Hence, heat absorbed by the cup = (9.8 cal/°C * 14.2°C) = 139.16 cal (ans)

(b) The mass of the solution is 100.0 gm, the specific heat of the solution is 1.03 cal/gm°C and the temperature rose from 22.3°C to 36.5°C.

Heat absorbed by the solution = (100.0 gm).(1.03 cal/gm °C).(36.5 – 22.3)°C = 1462.6 cal.

(c) The heat of neutralization is the sum of the heat absorbed by the solution and the Styrofoam cup. Thus, the heat of neutralization = (1462.6 + 139.16) cal = 1601.76 cal.

(d) Moles of acid used = (volume of acid added in L).(Molarity of the acid) =

(0.05 L)(2.0 mol/L) = 0.1 mol.

The molar heat of neutralization = heat of neutralization/ moles of acid =

1601.76 cal/ 0.1 mol = 16017.6 cal/mol = 16.0176 kcal/mol

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