Chapter 16
Study Guide - Answers
1.
The standard values that apply to the standard state enthalpy of a substance are 1 atmosphere & 25oC.
2.
Elemental substances have a standard enthalpy value of zero.
3.
What is the difference between H0 & H? ∆Ho represent the ∆H under standard conditions (25oC & 1atm)
4.
What is the difference between specific heat and heat capacity? Specific heat is specific to 1gram of a
substance.
5.
What is the relationship between qsur and qrxn? qrxn = -qsur
6.
Describe Hess’s law.
The enthalpy change for a reaction is equal to the enthalpy changes of each elementary step in a net reaction.
7.
What are the two common values used as the specific heat of water? 1 cal/g●oC & 4,184 J/g●oC
D
73kJ
56kJ A
Energy 
30kJ
E
C
B
Course of reaction 
8.
The reaction depicted by this diagram is exothermic / endothermic (circle one).
9.
How much energy is present in the activated complex of the reaction shown above?
73kJ .
10. The activation energy of this reaction is 17kJ .
11. The energy change (H) in the system is -26kJ .
12. The energy of the products in this reaction is 30kJ .
13. Phosphorus trichloride PCl3, is a compound used in the manufacture of pesticides and gasoline additives. How
much heat is required to raise the temperature of 158.7 g of PCl3 from 39.7oC to 89.9oC, if the specific heat is
0.874 J/goC?
q = 158.7g ● 0.874 J/goC ● 50.2oC
q = 6963J
Chapter 16
Study Guide - Answers
14. A piece of metal with a mass of 45.8-g is heated to 90.0oC and dropped into 40.0 mL of water at 21.0oC. The
final temperature of the system is 65oC. What is the specific heat of the metal?
First, we’ll find the q of the water (qsur) by using the mass of the water, specific heat of water (I’m going to
use joules) and the temperature change of the water (65oC – 21.0oC = 44oC)
qH2O = 40.0g ● 4.184 J/goC ● 44oC
qH2O = 7364 J
Now, since the water gained 7364 J, that heat must have been released by the hot metal. So the qMETAL may be
found by using the rule that qrxn = -qsur
qMETAL = - qH2O
qMETAL = -7364 J
Now, we’ll use this value with the mass of the metal, and temperature change of the metal (65oC – 90.0oC = 25oC [The metal is getting cooler in our system. It started at 90oC and ends up the same as the water at 65oC, so
it’s ∆T is negative)
-7364 J = 45.8g ● CMETAL ● -25oC
6.43 J/g●oC = CMETAL
15. Compute the heat released by the formation of 193g of ammonium bromide from ammonia and hydrogen
bromide.
NH3 + HBr  NH4Br ; H0= -188kJ.
193g NH4Br
x
1 mol NH4Br
97.943g NH4Br
x
188kJ
1 mol NH4Br
=
370kJ
=
2.74kJ
16. How much heat is generated by the reaction of 1.99g of Na2O2 with water?
2Na2O2 + 2H2O  4NaOH + O2 ; H0 = -215kJ.
1.99g Na2O2
x
1 mol Na2O2
77.978g Na2O2
x
215kJ
2 mol Na2O2