The balanced equation for the reaction is caco 3 s

The balanced equation for the reaction is:
CaCO
3
( s )
  
CaO( s )  CO
2
( g )
rxn f f 2 f 3
[ (CaO) (CO )] (CaCO )        H H H H
o o o o
198

CHAPTER 6: THERMOCHEMISTRY
rxn
[(1)( 635.6 kJ/mol) (1)( 393.5 kJ/mol)] (1)( 1206.9 kJ/mol) 177.8 kJ/mol         H
o
199

CHAPTER 6: THERMOCHEMISTRY
The enthalpy change calculated above is the enthalpy change if 1 mole of CO
2
is produced. The problem
asks for the enthalpy change if 66.8 g of CO
2
are produced. We need to use the molar mass of CO
2
as a
conversion factor.
2
2
2 2
1 mol CO 177.8 kJ
66.8 g CO
44.01 g CO 1 mol CO
   
2
2.70 10 kJ    H
6.61 Reaction   H   (kJ/mol)
S(rhombic)  O
2
( g )  SO
2
( g )  296.06
SO
2
( g )  S(monoclinic)  O
2
( g ) 296.36
S(rhombic)  S(monoclinic)
rxn
  0.30 kJ/mol H
o
Which is the more stable allotropic form of sulfur?
6.62 Strategy: Our goal is to calculate the enthalpy change for the formation of C
2
H
6
from is elements C and
H
2
. This reaction does not occur directly, however, so we must use an indirect route using the information
given in the three equations, which we will call equations (a), (b), and (c).
Solution: Here is the equation for the formation of C
2
H
6
from its elements.
2C(graphite)  3H
2
( g )
  
C
2
H
6
( g )
rxn
?   H
o
Looking at this reaction, we need two moles of graphite as a reactant. So, we multiply Equation (a) by two to
obtain:
(d) 2C(graphite)  2O
2
( g )
  
2CO
2
( g )
rxn
2( 393.5 kJ/mol) 787.0 kJ/mol      H
o
Next, we need three moles of H
2
as a reactant. So, we multiply Equation (b) by three to obtain:
(e) 3H
2
( g ) 
3
2
O
2
( g )
  
3H
2
O( l )
rxn
3( 285.8 kJ/mol) 857.4 kJ/mol      H
o
Last, we need one mole of C
2
H
6
as a product. Equation (c) has two moles of C
2
H
6
as a reactant, so we need
to reverse the equation and divide it by 2.
(f) 2CO
2
( g )  3H
2
O( l )
  
C
2
H
6
( g ) 
7
2
O
2
( g )
1
rxn
2
(3119.6 kJ/mol) 1559.8 kJ/mol    H
o
Adding Equations (d), (e), and (f) together, we have:
Reaction   H   (kJ/mol)
(d) 2C(graphite)  2O
2
( g )
  
2CO
2
( g )  787.0
(e) 3H
2
( g ) 
3
2
O
2
( g )
  
3H
2
O( l )  857.4
(f) 2CO
2
( g )  3H
2
O( l )
  
C
2
H
6
( g ) 
7
2
O
2
( g ) 1559.8
2C(graphite)  3H
2
( g )
  
C
2
H
6
( g )  H    84.6 kJ/mol
200

CHAPTER 6: THERMOCHEMISTRY
6.63 Reaction   H   (kJ/mol)
CO
2
( g )  2H
2
O( l )  CH
3
OH( l ) 
3
2
O
2
( g ) 726.4
C(graphite)  O
2
( g )  CO
2
( g )  393.5
2H
2
( g )  O
2
( g )  2H
2
O( l ) 2(  285.8)
C(graphite)  2H
2
( g ) 
1
2
O
2
( g )  CH
3
OH( l )
rxn
  238.7 kJ/mol H
o

We have just calculated an enthalpy at standard conditions, which we abbreviate
rxn
 H
o
. In this case, the
reaction in question was for the formation of one mole of CH
3
OH from its elements in their standard state.
Therefore, the
rxn
 H
o
that we calculated is also, by definition, the standard heat of formation
f
 H
o
of CH
3
OH
(  238.7 kJ/mol ).