Question

CH
6.20 The equilibrium constant for a reaction is 10. What will be the value
of AGⓇ? R = 8.314 JK-' mol'. T = 300 K.
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Answer 1

Explanation:

CH

6.20 The equilibrium constant for a reaction is 10. What will be the value

of AGⓇ? R = 8.314 JK-' mol'. T = 300 K.

Answer 2

Answer:

ΔG

0

=−2.303RTlogK=−2.303×8.314×300log10=−5.744kJ/mol

Explanation:

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