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)).`

Options

A. `2.846 xx 10^(-3)`
B. `6.481 xx 10^(-3)`
C. `1.856 xx 10^(-3)`
D. `1.648 xx 10^(-3)`

Answer

D

Explanation

Given that `Delta U - Delta H = 1200` cal where `Delta U = q_(v)` and `Delta H = q_(p)`
`Delta nRT = -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)`