NAME INSTRUCTOR: DATE: SECTION/GROUP PRE-LABORATORY QUESTIONS THERMOCHEMISTRY: A

Question

NAME INSTRUCTOR: DATE: SECTION/GROUP PRE-LABORATORY QUESTIONS THERMOCHEMISTRY: ACID-BASE NEUTRALIZATION I. Consider this coffee cup calorimetry experiment: a. 100 mL of water at 86.0 C was poured into 100 ml. of water at 23.5°C in a coffee cup Calculate the heat capacity of calorimeter, and the temperature of the contents of the calorimeter equilibrated to 53.5 °C before beginning to fall to room temperature. the calorimeter. (The density and specific heat of water or any aqueous solution can be assumed to be 1.00 g/mL. and 4.184 J/g. "C, respectively) 100 mL of 2.00M acetic acid was added to 100 mL of 2.00M NaOH at 23.5°C and the temperature of the solution equilibrated to 36.3°C. Calculate an enthalpy of neutralization for acetic acid, using the calculated heat capacity of the coffee cup. b. Laboratory Thermochemistry: Acid-Base Nestralization

Explanation / Answer

(2)

The main equation given to us is:

2C(s) + H2(g) ---> C2H2(g)

Data equations given:

C2H2(g) + 2.5O2(g) ---> 2CO2(g) + H2O(l) ---(1)

C(s) + O2(g) ---> CO2(g) ---(2)

H2(g) + 0.5O2(g) ---> H2O(l) ---(3)

In order to get this equation, we multiply second equation by 2, then add the third equation and then subtract the first equation.

So according to Hess' law, for the final equation we have:

dHrxn = 2*dHrxn2 + dHrxn3 - dHrxn1

Putting values:

dHrxn = 2*(-393.5) + (-285.8) - (2599) = -3671.8kJ

So, enthalpy of formation = -3671.8 kJ/mol

Hope this helps !

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.