1.

For a system at equilibrium which of the following are correct

Answer»

`((d"In"K_p)/(dP))_r=(-d)/(dP)((DeltaG)/(RT))_r`
`log K_p =1/(2.303R)(DeltaS-(DeltaH)/T)`
on increasing the TEMPERATURE of an endothermicreaction, the equilibrium shift in forward direction because Kincreases
on increasing the temperature of an ENDOTHERMIC reaction the equilibrium shifts in forward direction because Q decreases

Solution :`DeltaG=RT"In" K_p`
`therefore ((d"In"K_p)/(dP))_T=-d/(dP)((DeltaG)/(RT))_T`
Hence (A) is the CORRECT option.
Also, `DeltaG=DeltaH-TDeltaS=-2.303 RT log K`
`therefore log K=1/(2.303)(DeltaS-(DeltaH)/T)`
Hence (B) is correct option.
In endothermic reaction changing temp. K changes and not Q .
Hence ( C ) is the correct option.


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