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Chemistry
Chemistry
Problem
The enthalpy change for the following reaction at
25
∘
C
25^\circ\text{C}
2
5
∘
C
are given below :
1
2
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
OH
(
g
)
;
Δ
H
=
−
10.06
kcal
…
(
1
)
\frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \longrightarrow \text{OH}(g); \Delta H = -10.06 \text{ kcal} \dots(1)
2
1
H
2
(
g
)
+
2
1
O
2
(
g
)
⟶
OH
(
g
)
;
Δ
H
=
−
10.06
kcal
…
(
1
)
H
2
(
g
)
⟶
2
H
(
g
)
;
Δ
H
=
104.18
kcal
…
(
2
)
\text{H}_2(g) \longrightarrow 2\text{H}(g); \Delta H = 104.18 \text{ kcal} \dots(2)
H
2
(
g
)
⟶
2
H
(
g
)
;
Δ
H
=
104.18
kcal
…
(
2
)
O
2
(
g
)
⟶
2
O
(
g
)
;
Δ
H
=
118.32
kcal
…
(
3
)
\text{O}_2(g) \longrightarrow 2\text{O}(g); \Delta H = 118.32 \text{ kcal} \dots(3)
O
2
(
g
)
⟶
2
O
(
g
)
;
Δ
H
=
118.32
kcal
…
(
3
)
Calculate OH bond energy in hydroxyl radical.
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