The enthapies of formation of `N_2O` and NO are 28 and 90 kj `mol^(-1)` respectively. The enthalpy of the reaction, `2N_2O(g)+O_2(g) rarr 4NO(g)` is equal to

Options

A. 8 kJ
B. 88 kJ
C. `-16 kJ`
D. `304 kJ`

Answer

D

Explanation

`N_2+1/2 O_2 rarr N_2O, Delta H= 90 kj ….(i) `
`1/2 N_2 + 1/2 O_2 rarr N_2O, Delta H = 90 kj …… (ii)`
By `[4xx ii]-[2 xx (i) ]`
`2N_2O+O_2 rarr 4 NO , Delta H = 304 kJ `