The free energy of formation of NO is `78 kJ mol^(-1)` at the temperature of an authomobile engine `(1000 K)`. What is the equilibrium constant for this reaction at `1000 K`?
`1/2 N_(2)(g)+1/2 O_(2)(g) hArr NO(g)`

Options

A. `8.4 xx 10^(-5)`
B. `7.1 xx 10^(-9)`
C. `4.2 xx 10^(-10)`
D. `1.7 xx 10^(-19)`

Answer

A

Explanation

`A_((g)) + B_((g)) rarr C_((g)) , Delta E= -5` cal
`Delta n_(g)=1 - (1+1)= -1`
`Delta H= Delta E + Delta n RT= -5-1 xx 2.0 xx 298 = -601`
`Delta G= Delta H- T Delta S= -601 - 298 xx (-10) = 2379` cal