For the part circled in white, I don't understand when simplifying this why we don't divide the coefficient in front of octane by 2. I thought it would be 1/2 Since we divided the other ones by 2.

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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For the part circled in white, I don't understand when simplifying this why we don't divide the coefficient in front of octane by 2. I thought it would be 1/2 Since we divided the other ones by 2.

In the second part, calculate AHn from the
AH's of the reactants and products.
(2)
AH xn
Txn
ΔΗ's
ΔΗ Σn,ΔΗ (products) - Σn,ΔΗ?(reactants)
AHTXN
In the third part, convert from kilojoules of
energy to moles of octane using the conver-
sion factor found in step 2, and then convert
from moles of octane to mass of octane using
(3)
kJ
mol CgH18
g C8H18
kg C3H1
114.22 g C8H18
1 kg
the molar mass.
Conversion factor to be
mol C3H18
1000 g
determined from steps 1 and 2
RELATIONSHIPS USED
molar mass C3H18 = 114.22 g/mol
1 kg = 1000 g
SOLVE Begin by writing the balanced equa-
JLUTION STEP 1
tion for the combustion of octane. For con-
C3H18(1) +
25
O2(8) 8 CO2(8) + 9 H2O(g)
venience, do not clear the 25/2 fraction in
order to keep the coefficient on octane as 1.
SOLUTION STEP 2
Reactant or product
AH; (kJ/mol, from Appendix IIB)
CgH18(/)
-250.1
O2(g)
0.0
Look up (in Appendix IIB) the standard
enthalpy of formation for each reactant and
product and then calculate AHxn:
CO2(g)
-393.5
H20(g)
-241.8
-Continued on the nex
Transcribed Image Text:In the second part, calculate AHn from the AH's of the reactants and products. (2) AH xn Txn ΔΗ's ΔΗ Σn,ΔΗ (products) - Σn,ΔΗ?(reactants) AHTXN In the third part, convert from kilojoules of energy to moles of octane using the conver- sion factor found in step 2, and then convert from moles of octane to mass of octane using (3) kJ mol CgH18 g C8H18 kg C3H1 114.22 g C8H18 1 kg the molar mass. Conversion factor to be mol C3H18 1000 g determined from steps 1 and 2 RELATIONSHIPS USED molar mass C3H18 = 114.22 g/mol 1 kg = 1000 g SOLVE Begin by writing the balanced equa- JLUTION STEP 1 tion for the combustion of octane. For con- C3H18(1) + 25 O2(8) 8 CO2(8) + 9 H2O(g) venience, do not clear the 25/2 fraction in order to keep the coefficient on octane as 1. SOLUTION STEP 2 Reactant or product AH; (kJ/mol, from Appendix IIB) CgH18(/) -250.1 O2(g) 0.0 Look up (in Appendix IIB) the standard enthalpy of formation for each reactant and product and then calculate AHxn: CO2(g) -393.5 H20(g) -241.8 -Continued on the nex
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