when 2 mole of H2 and 1 mole of O2 at 100°C and 1 torr pressure react to produce 2 mole of gaseous water, 484.5 KJ of energy are evolved. What are the values of ∆H and ∆E for the production of H2O.?
Answers
Explanation:
tefan V.
Feb 27, 2018
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
l
)
Δ
H
=
−
276 kJ
Explanation:
You know that when
2
moles of hydrogen gas react with
1
mole of oxygen gas, you get
2
moles of water and
572 kJ
of heat are evolved.
2
H
2
(
g
)
+
O
2
(
g
)
→
2
H
2
O
(
l
)
Δ
H
=
−
572 kJ
Don't forget that the enthalpy change of reaction must be negative here to illustrate the fact that heat if being given off by the reaction.
Now, in order for this reaction to produce
1
mole of water, all the coefficients of the chemical equation must be halved.
(
1
2
⋅
2
)
H
2
(
g
)
+
1
2
O
2
(
g
)
→
(
1
2
⋅
2
)
H
2
O
(
l
)
This will get you
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
l
)
Now, the enthalpy change for this reaction will be half the value of the enthalpy change for the reaction that produced
2
moles of water.
Δ
H
1 mole H
2
O
=
1
2
⋅
Δ
H
2 moles H
2
O
Δ
H
1 mole H
−
2
O
=
−
572 kJ
2
=
−
286 kJ
This means that the thermochemical equation that describes the formation of
1
mole of water looks like this
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
l
)
Δ
H
=
−
276 kJ
To write the thermochemical equation that describes the decomposition of
1
mole of water into hydrogen gas and oxygen gas, you need to reverse the chemical equation
H
2
O
(
l
)
→
H
2
(
g
)
+
1
2
O
2
(
g
)
and change the sign of the enthalpy change of reaction.
Δ
H
reverse
=
−
Δ
H
forward
This means that the thermochemical equation will look like this
H
2
O
(
l
)
→
H
2
(
g
)
+
1
2
O
2
(
g
)
Δ
H
=
+
276 kJ
This means that when
1
mole of water undergoes decomposition,
276 kJ
of heat are being absorbed