The equilibrium constant `K_(p)`, for the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` is `1.6xx10^(-4)` at `400^(@)C`. What will be the equilibrium constant at `500^(@)C` if the heat of reaction in this temperature range is `-25.14` kcal?

Options

A. `1.231xx10^(-4) ("atm")^(-2)`
B. `1.876xx10^(7) ("atm")^(-2)`
C. `1.462xx10^(-5) ("atm")^(-2)`
D. `3.462xx10^(-5) ("atm")^(-2)`

Answer

C

Explanation

Equilibrium constants at different temperature and heat of the reaction are related by the equation.
In `(K_(p1))/(K_(p1))=(Delta H^(@))/(R ) [(1)/(T^(2))-(1)/(T^(2))]`
`2.303" log "(K_(p2))/(K_(p1)) =(Delta H^(@))/(R ) [(T_(2)-T_(1))/(T_(1)T_(2))]`
log `K_(p2) =(-25140)/(2.303xx2) [(773-673)/(773xx673)]+ log (1.6xx10^(-4))`
`log K_(p2) =-4.835`
`therefore K_(p2) =1.462xx10^(-5)" (atm)"^(-2)`