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The heat of reaction for an endothermic ...

The heat of reaction for an endothermic reaction at constant volume in equilibrium is 1200 cal more than al constant pressure at 300K, then

A

`Deltan_((g))= -2`

B

`(K_(P))/(K_(c))=1.648 xx 10^(-3)`

C

`Deltan_((g))= +2`

D

`(K_(c))/(K_(P))=1.648 xx 10^(-3)`

Text Solution

Verified by Experts

The correct Answer is:
A, B
`DeltaU=DeltaH+1200`
`DeltaU-DeltaH=1200=Delta n_((g))RT`
`Delta n_((g))=(-1200)/(2 xx 300)= -2 implies (Kp)/(Kc)=(RT)^(-2)=(0.0821 xx 300)^(-2)=1.648 xx 10^(-3)`
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