1.

Consider the reaction : 4NO_(2)(g) + O_(2)(g) rarr 2N_(2)O_(5)(g), Delta_(f)H = -111 kJ If N_(2)O_(5)(s) is fromed instead of N_(2)O_(5)(g) in the above reaction, the Delta_(f)H is value will be (given, Delta H of sublimation for N_(2)O_(5) is 54 kJ mol^(-1)).

Answer»

`+ 54 kJ`
`+ 219 kJ`
`- 219 kJ`
`- 165 kJ`

Solution :`4NO_(2) + O_(2) rarr 2N_(2)O_(5) (g) Delta H = -111 kJ`
`2N_(2)O_(5) (s) rarr 2N_(2)O_(5) (g) Delta H = 54 xx 2 kJ`
SUBTRACTING
`4NO_(2) + O_(2) rarr 2N_(2)O_(5) (s) Delta H = - 219 kJ`


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