The Richard Stockton College of New Jersey
Chemistry Program, School of Natural Sciences and Mathematics
PO Box 195, Pomoma, NJ
CHEM 3410: Physical Chemistry I — Fall 2008
Homework 4— Solutions
1. (a) The balanced chemical reaction would look like this:
Fe2O3+ 2 Al →2 Fe + Al2O3
The enthalpy change for this reaction at 298 K can be calculated using standard enthalpies
of formation of the products and reactants:
∆H◦
rxn =∆H◦
f(Al2O3) + 2∆H◦
f(F e)−∆H◦
f(F e2O3) + 2∆H◦
f(Al)
The standard enthalpies of formation of elements are zero, so using available thermochemical
data:
∆H◦
rxn =−851.5 kJ/mole
(b) To answer this part of the question, you needed more information. From the heat capacity,
you could calculate the heat necessary to heat the iron from room temperature to its melting
point. However, I neglected to give you the enthalpy of melting of iron, nor did I give you
data on the solid-solid phase transformation that occurs below the melting point (see the
latest batch of in-class problems).
2. (a) To find the final temperature, we need to find out how much heat is required to raise the
temperature in each phase and each phase temperature. The best way to do this is in steps.
i. First ask if there is enough heat coming into the system to raise the temperature from
the initial temperature (298 K) to the first transition temperature (α→β, 993 K). The
heat required to raise the temperature this much is:
q=nZ992
298
Cα
pdT = 2 Z992
298 21.6159 J
mole ·KdT = (2)(21.6159)(993 −298) = 30046 J
We have a total of 100698 J of heat available. Therefore we have enough to raise the
temperature to the first transition (from αto β) and have 100698 −30046 = 70652 J of
heat remaining.
ii. Now do we have enough heat available to complete the transition from the αto the β
phase:
∆Hα→β
trans = (2 moles)(2010 J /mole) = 4020 J
We have 70652 J remaining from (i), so we can completely transform to the βphase,
leaving 70652 −4020 = 66632 J of heat remaining.
iii. Now can we raise the temperature from 993 K to the transition temperature for βto γ
at 1373K?
q=nZ1373
993
Cβ
pdT = 2 Z1373
993 34.9028 J
mole ·KdT = (2)(34.9028)(1373−993) = 26526 J
Yes we can! This leaves us with 66632 −26526 = 40106 J leftover. So we can continue!
iv. Now do we have enough heat available to complete the transition from the βto the γ
phase:
∆Hβ→γ
trans = (2 moles)(2300 J /mole) = 4600 J
This leaves us 40106 −4600 = 35506 J leftover.
