CHEM 103 MODULE 3 RATED
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Question 1 10 / 10 pts
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A reaction between HCl and NaOH is being studied in a styrofoam coffee cup with a lid and the heat given off is measured by means of a thermometer immersed in the reaction mixture. Enter the correct thermochemistry term to describe the item listed.
1. The type of thermochemical process
2. The calorimeter and mixture of HCl + NaOH Your Answer:
1. Heat given off = Exothermic process
2. With lid = closed system
Question 2 10/ 10 pts
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1. Show the calculation of the final temperature of the mixture when a 22.8 gram sample of water at 74.6oC is added to a 14.3 gram sample of water at 24.3oC in a coffee cup calorimeter.
c (water) = 4.184 J/g oC
2. Show the calculation of the energy involved in freezing 54.3 grams of water at 0oC if the Heat of Fusion for water is 0.334 kJ/g Your Answer:
1. - (mwarn H2O x cwarn H2O x Δtwarn H2O) = (mcool H2O x ccool H2O x Δtcool H2O)
- [22.8 g x 4.184 J/g oC x (Tmix - 74.6oC)] = [14.3 g x 4.184 J/g oC x (Tmix -
24.3oC)]
- [95.3952 J/oC x (Tmix - 74.6oC)] = [59.8312 J/oC x (Tmix - 24.3oC)] Tmix = 55.2oC
2. ql↔s = m x ΔHfusion = 54.3 g x 0.334 kJ/g = 18.14 kJ (since heat is removed ) = - 18.14 kJ
Question 3 10 / 10 pts
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Show the calculation of the ΔH of the reaction if 20.7 g of S is reacted with excess O2 to yield 255.6 kJ and sulfur trioxide by the following reaction equation. Report your answer to 4 significant figures.
2 S (s) + 3 O2 (g) → 2 SO3 ( g )
Your Answer:
2 S (s) + 3 O2 (g) → 2 SO3 ( g ) ΔHrx is for 2 mole of S reaction uses 20.7 g S = 20.7/32.07 = 0.6455 mole S
255.6 = ΔHrx x new moles / original moles
-255.6 = ΔHrx x 0.6455 mole S / 2 mole S = -791.9 kJ
Question 4 10 / 10 pts You may find the following resources helpful:
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Show the calculation of the heat of reaction (ΔHrxn) for the reaction : P4 (s) +
10 Cl2 (g) → 4 PCl5 ( s )
by using the following thermochemical data:
P4 (s) + 6 Cl2 (g) → 4 PCl3 (s) ΔH = - 1277.2 kJ
PCl3 (l) + Cl2 (g) → PCl5 (s) ΔH = - 123.8 kJ
Your Answer:
P4 (s) + 6 Cl2 (g) → 4 PCl3 (s) ΔH = - 1277.2 kJ
4 (PCl3 (l) + Cl2 (g) → PCl5 (s) ΔH = - 123.8 kJ)
P4 (s) + 10 Cl2 (g) → 4 PCl5 (s) ΔHrxn = - 1772.4 kJ
ΔHrxn = -1277.2 + 4 (-123.8) = - 1772.4
Question 5 10 / 10 pts You may find the following resources helpful:
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Show the calculation of the heat of reaction (ΔHrxn) for the reaction : 2 CH4 (g)
+ 3 O2 (g) → 2 CO (g) + 4 H2O (l) by using the following ΔHf0 data:
ΔHf0 CH4 (g) = -74.6 kJ/mole, ΔHf0 CO (g) = -110.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole Your Answer:
2 CH4 (g) + 3 O2 (g) → 2 CO (g) + 4 H2O (l)
ΔHf0 CH4 (g) = -74.6 kJ/mole, ΔHf0 CO (g) = -110.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole
ΔHrxn = 2(+74.6) + 3(0) + 2(-110.5) + 4(-285.8) = - 1215.0 kJ/mole [Show Less]