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The heat of reaction at constant volume ...

The heat of reaction at constant volume for an endothermic reaction in equilibrium is 1200 cal more than at constant pressure at:300 K, Calculate the ratio of equilibrium constants `K_(p)` (atm) and `K _(c) (mol L^(-1)).`

A

`2.846 xx 10 ^(-3)`

B

`6.481 xx 10 ^(-3)`

C

`1.856 xx 10 ^(-3)`

D

`1.648 xx 10 ^(-3)`

Text Solution

Verified by Experts

The correct Answer is:
D
Given that `Delta U - Delta H = 1200 cal ` where `Delta U = q _(v) and Delta H = q _(P)`
`Delta n RT =- 1200 or Delta n = (- 1200)/(2 xx 300) =- 2 (R = 2 cal)`
Now using `K _(P) = K _(c) (RT) ^( Delta n) , (K _(P))/( K _(c)) = (0.0821 xx 300) ^(-2) = 1 648 xx 10 ^(-3)`
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