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For a reaction whose standrad enthalpy c...

For a reaction whose standrad enthalpy change is -100 kJ, what final temperature is needed to double the equilibrium constant from its value at 298 K?

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`"log"K_(2)/K_(1)=(DeltaH^(@))/(2.303R)[(T_(2)-T_(1))/(T_(1)*T_(2))]`
`log2=(-100000)/(2.3xx8.3)[(T_(2)-298)/(298T_(2))]`
`0.301=(-100000)/(19.09)((T_(2)-298))/((298T_(2)))`
Solving for `T_(2)`, we get `T_(2)~~293K`
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