At 100°C and 1 atm 2 mol of H2 gas and 1 mol O2 gas react to form 2 mol H2Ogas releasing 484 KJ. 1) find the enthalpy change of the reaction. H2 (g) + 1/2 O2 (g) → H2O(g) (reaction 1) 2) At 100°C if the entropy of H2(g), O2(g), H2O(g) are 130 Jmol-1K-1, 205Jmol-1K-1, 190 Jmol-11K-1 respectively , Calculte the entropy change of reaction 1 3) Find ΔG of reaction 1 at 100°C 4) Will reaction 1 takeplace spontaneously at 100°C. Give reason Please find this! I need it!
Answers
Answer:
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