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Chemistry
Chemistry
Problem
At
298
K
298\,\text{K}
298
K
the free energy change for a reaction is given by
Δ
G
(
kJ
)
=
5.7
+
2.48
ln
Q
\Delta G\ (\text{kJ}) = 5.7 + 2.48 \ln Q
Δ
G
(
kJ
)
=
5.7
+
2.48
ln
Q
, where
Q
Q
Q
is the reaction quotient. Calculate the value of
Q
Q
Q
at which the reaction is at equilibrium at this temperature.
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