CHEM 103 MUDULE 1,2,3,4,5& 6 EXAM QUESTIONS AND ANSWERS VERIFIED 100% CORRECT| PORTAGE LEARNING|

Portage Learning / Chem 103 / module 1-6 Exam with Answers




CHEM 103 MUDULE 1,2,3,4,5& 6 EXAM
QUESTIONS AND ANSWERS VERIFIED 100%
CORRECT| PORTAGE LEARNING|


MODULE 3 EXAM
Question 1

Click this link to access the Periodic Table. This may be helpful throughout the exam.


A reaction between HCl and NaOH is being studied in a styrofoam coffee cup with NO
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 amount of heat released in the reaction of HCl with NaOH


1. Heat given off = Exothermic process


2. The amount of heat released = Heat of reaction


Question 2
Click this link to access the Periodic Table. This may be helpful throughout the exam.


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

,Portage Learning / Chem 103 / module 1-6 Exam with Answers




2. Show the calculation of the energy involved in freezing 54.3 grams of water at 0 oC if
the Heat of Fusion for water is 0.334 kJ/g


1. - (mwarn H2O x cwarn H2O x ∆twarn H2O) = (mcool H2O x ccool H2O x ∆tcool 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 ∆Hfusion = 54.3 g x 0.334 kJ/g = 18.14 kJ (since heat is removed) = -
18.14 kJ


Question 3
Click this link to access the Periodic Table. This may be helpful throughout the exam.


Show the calculation of the amount of heat involved if 35.6 g of H2S is reacted with
excess O2 to yield sulfur trioxide and water by the following reaction equation.
Report your answer to 4 significant figures.

2 H2S (g) + 3 O2 (g) → 2 SO2 (g) + 2 H2O (g) ΔH = - 1124 kJ




1 mol H2s = 34.1 g of H2S = 603.2 kj
(35.6g/34.1 g) x -603.2 kJ = (1.0439) x (-603.2 kJ) = -629. 7 kj


2 H2S (g) + 3 O2 (g) → 2 SO2 (g) + 2 H2O (g) ΔH = - 1124 kJ

ΔHrx is for 2 mole of H2S

reaction uses 35.6 g of H2S = 35.6/34.086 = 1.044 mole of H2S q = ΔHrx x
new moles / original moles
q = -1124 kJ x 1.044 mole of H2S / 2 mole H2S = 586.7 given off

,Portage Learning / Chem 103 / module 1-6 Exam with Answers


Question 4
Click this link to access the Periodic Table. This may be helpful throughout the exam.
Show the calculation of the heat of reaction (ΔHrxn) for the reaction:

3 C (graphite) + 4 H2 (g) → C3H8 (g) by using the
following thermochemical data:


C (graphite) + O2 (g) → CO2 (g) kJ ΔH = - 393.51

ΔH = - 571.66
2 H2 (s) + O2 (g) → 2 H2O(l) kJ

ΔH = - 2220.0 kJ
C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O(l) Your Answer:

3 (C (graphite) + O2 (g) → CO2 (g) 393.51 kJ) ΔH = -

2 (2 H2 (s) + O2 (g) → 2 H2O(l) - ΔH =
571.66 kJ)
3 CO2 (g) + 4 H2O(l) → C3H8 (g) + 5 O2 (g) ΔH = +
2220.0 kJ

3 C (graphite) + 4 H2 (g) → C3H8 (g) ΔHrxn =
- 103.85 kJ
ΔHrxn = 3(- 393.51) + 2(- 571.66) + 2220.0 = - 103.85

Question 5
Click this link to access the Periodic Table. This may be helpful throughout the exam.


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:

ΔH f0 CH 4 (g) = -74.6 kJ/mole, ΔHf 0 CO (g) = -110.5 kJ/mole, ΔHf 0 H2 O (l) = -285.8 kJ/mole

2 CH4 (g) + 3 O2 (g) → 2 CO (g) + 4 H2O (l)

, Portage Learning / Chem 103 / module 1-6 Exam with Answers



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



Question 6
Click this link to access the Periodic Table. This may be helpful throughout the exam.


Show the calculation of the new temperature of a gas sample has an original volume of
740 ml when collected at 710 mm and 35oC when the volume becomes 460 ml at
1.20 atm.
P1V1/T1 = P2V2/T2
T2 = 1.2 x 460x.934 x 740 = 245.98 kelvin = -27.17oC

(Pi x Vi ) / Ti = (Pf x Vf ) / Tf


740 ml/1000 = 0.740 liters = Vi 710
mm/760 = 0.934 atm = Pi
460 ml/1000 = 0.460 liters = Vf
1.20 atm = Pf
35oC + 273 = 308oK = Ti


(0.934) x (0.740) / 308 = (1.20) x (0.460) / Tf


Tf = 246 oK




Question 7
Click this link to access the Periodic Table. This may be helpful throughout the exam.


Show the calculation of the volume occupied by a gas sample containing 0.632 mole
collected at 710 mm and 35oC.