Answer is: -1.24 x 103 kJ/mol the
The answer is right above so don't worry about reading all these.
reason why ---- The heat capacity of the calorimeter is
c
=
23.3
k
J
/
o
C
,
and the change in temperature during the reaction is
Δ
T
=
76.0
−
35.0
o
C
=
41
o
C
.
The heat change during this combustion is:
Q
=
c
Δ
T
=
(
23.3
k
J
/
o
C
)
(
41
o
C
)
=
955.3
k
J
.
We burned 35.6 g of ethanol, whose molar mass is given as 46.07 g/mol. How many moles are these? We evaluate:
35.6
g
46.07
g
/
m
o
l
=
0.733
m
o
l
.
The enthalpy of reaction is therefore:
Δ
H
=
955.3
k
J
0.773
m
o
l
=
1236
k
J
/
m
o
l
.
Since this is an exothermic (heat-releasing) reaction, we conventionally introduce a negative sign and write:
Picture and explanation: C2H5OH(l) + O2(g) --> CO2(g) + H2O(g) ?H_ rxn = ?
-1.24 * 103 kJ/mol
+1.24 * 103 kJ/mol
-8.09 * 103 kJ/mol
-9.55 * 103 kJ/mol
+9.55 * 103 kJ/mol
Things you need to know: Enthalpy of Reaction
The enthalpy of a chemical reaction is the amount of heat released or absorbed when the reactants' bonds are broken and the products' bonds are formed. Exothermic reactions give off heat and are assigned a negative sign by convention, while endothermic ones absorb heat are carry a positive
