Byju's Answer
Standard XII
Chemistry
Heat Capacity at Constant Volume
If K =1.19 × ...
Question
If K=1.19×10-4 at 623K. Find 🔼G
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Solution
ΔG = - RT lnK (remember this equation)
= - 8.314 * 623 * ln (1.9 * 10
-4
) = -32455.007 J = -322.455007 KJ.
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