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Chemistry
Chemistry
Problem
Show that
2.303
log
10
K
=
−
Δ
H
∘
R
T
+
Δ
S
∘
R
2.303 \log_{10} K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}
2.303
lo
g
10
K
=
−
R
T
Δ
H
∘
+
R
Δ
S
∘
; where
K
K
K
is equilibrium constant and
Δ
H
∘
\Delta H^\circ
Δ
H
∘
and
Δ
S
∘
\Delta S^\circ
Δ
S
∘
are standard enthalpy and standard entropy change at temperature
T
T
T
.
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