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| | Message 1 of 21 in Discussion |
| From: Abalo3 (Original Message) | Sent: 1/3/2008 9:15 PM |
I am stuck and need some help. Please Thank you
Questions
(f) calculate the calorimeter's heat absorption as Qcal = Ccal * delta T (Joules):
(g) record the true heat of reaction as Qreaction = Qobserved + Qcal (Joules):
2. Convert the total heat of dissolution to a molar enthalpy:
(a) moles of NH4Cl in 5g (MW = 53.492)
(b) enthalpy of the reaction, per mole of NH4Cl (J/mol):
3. Write out the reaction NH4Cl(s) -> NH3(g) + HCl(g) as a series of steps which include the reactions observed in Procedures 2 and 3.
Use the following known (previously determined) enthalpies of reaction as additional information:
Change of state of NH3(g) -> NH3(aq) (delta H = -34640 J/mol)
Change of state of HCl(g) -> HCl(aq) (delta H = -75140 J/mol)
Combine the above data and the experimental data values you determined in this lab as needed to allow calculation of the enthalpy change of the desired final reaction.
Be sure to show how the reaction steps must proceed so that "delta H" for the desired reaction can be calculated. And be careful to use the correct positive or negative enthalpy values depending on the DIRECTION of the reactions that you add together.
In your final answer specify whether the enthalpy change occurring in the combined final reaction causes heat to be released or absorbed.
Answers 1. a. volume of water in the calorimeter was 25.00 ml
b. Mass of water in the calorimeter (assuming a density of 1.0 g/mL for water) (g): =Volume * density = 25ml* 1.0 g/ml = 25 g
c. Total mass in the calorimeter at the end of the reaction: = 25g + 5g =30 grams
d. Temperature change from the reaction =21.00 degrees C - 10.09 degrees C = 10.91 degrees C By adding Ammonium Chloride to the water , the temperature started increasing from 10.09 degrees C to 21.00 degrees C. Such reaction in temperature is proved to be endothermic because the change of heat is positive meaning no heat flows.
e. The observed heat of reaction in joules =30g * 4.184 j/g-Deg C * 10.91Deg C =1369.42 joules
f. The calorimeter's heat absorption =
f. Calorimeter's heat absorption
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