Question #658ea

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Answer 1 · 2015-01-22T22:30:21

$K=3xx10^5$

$DeltaG=-RTlnK$

$lnK=-(DeltaG)/(RT)$

$lnK=-((-32.5xx10^3)/(8.31xx310))$

$lnK=12.61$

$K=3xx10^5$

Strictly speaking $DeltaG^0$ should be under standard conditions. This is at 310K.