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For H = -28 kcal and H = -36.8 kcal If 1 mole of was mixed with 4 mole of HCl, 0.44…
CHEMISTRY · Subjective · A Level · Chemistry · Energetics · Energetics
Problem
For
Au
(
OH
)
3
+
4
HCl
⟶
HAuCl
4
+
3
H
2
O
;
\text{Au}(\text{OH})_3 + 4\text{HCl} \longrightarrow \text{HAuCl}_4 + 3\text{H}_2\text{O};
Au
(
OH
)
3
+
4
HCl
⟶
HAuCl
4
+
3
H
2
O
;
Δ
H
=
−
28
kcal
\Delta H = -28 \text{ kcal}
Δ
H
=
−
28
kcal
and
Au
(
OH
)
3
+
4
HBr
⟶
HAuBr
4
+
3
H
2
O
;
\text{Au}(\text{OH})_3 + 4\text{HBr} \longrightarrow \text{HAuBr}_4 + 3\text{H}_2\text{O};
Au
(
OH
)
3
+
4
HBr
⟶
HAuBr
4
+
3
H
2
O
;
Δ
H
=
−
36.8
kcal
\Delta H = -36.8 \text{ kcal}
Δ
H
=
−
36.8
kcal
If 1 mole of
HAuBr
4
\text{HAuBr}_4
HAuBr
4
was mixed with 4 mole of HCl, 0.44 kcal heat was absorbed. The percentage conversion of
HAuBr
4
\text{HAuBr}_4
HAuBr
4
to
HAuCl
4
\text{HAuCl}_4
HAuCl
4
is .....
Related theory
Hess's Law
Thermodynamics
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