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Chemistry
Chemistry
Problem
Given :
HCl
(
g
)
+
40
H
2
O
⟶
HCl
(
40
H
2
O
)
;
Δ
H
=
−
73.0
kJ
\text{HCl}_{(g)} + 40\text{H}_2\text{O} \longrightarrow \text{HCl}(40\text{H}_2\text{O}); \Delta H = -73.0 \text{ kJ}
HCl
(
g
)
+
40
H
2
O
⟶
HCl
(
40
H
2
O
)
;
Δ
H
=
−
73.0
kJ
HCl
(
g
)
+
10
H
2
O
⟶
HCl
(
10
H
2
O
)
;
Δ
H
=
−
69.0
kJ
\text{HCl}_{(g)} + 10\text{H}_2\text{O} \longrightarrow \text{HCl}(10\text{H}_2\text{O}); \Delta H = -69.0 \text{ kJ}
HCl
(
g
)
+
10
H
2
O
⟶
HCl
(
10
H
2
O
)
;
Δ
H
=
−
69.0
kJ
then integral heat of dilution for two moles of
HCl
(
10
H
2
O
)
\text{HCl}(10\text{H}_2\text{O})
HCl
(
10
H
2
O
)
to
HCl
(
40
H
2
O
)
\text{HCl}(40\text{H}_2\text{O})
HCl
(
40
H
2
O
)
is : a. -4.5 kJ b. -3.5 kJ c. -8.0 kJ d. -2.0 kJ
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