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Entropy
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is equilibrium depended on temperature
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yes
I mean no* sorry
logK=−ΔG0RT .
0=divided by
Equilibrium is definitely affected by temperature. If it wasn't, why is it in the equation?
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equilibrium is affected by a few factors temperature and pressure are 2
Temperature - how K changes with temperature depends on the type of reaction involved, i.e. whether or not it is exothermic or endothermic. If the reaction is exothermic, an increase in temperature will cause the reaction to be less product-favored at the higher temperatures (greater value in denominator of equilibrium expression), therefore value of K decreases. Whereas, for endothermic reactions the reverse is true.
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