In the following reaction: 4NO_(2)(g) + O_(2)(g) to 2N_(2)O_(5)(g),DeltaH = - 110 kJ, if N_(2)o_(5)(s) is formed instead of N_(2)O_(5)(g) in the reaction, the enthalpy change (in kJ) would be (enthalpy of sublimation of N_(2)O_(5)(s) is + 53 kJ mol^(-1)).

In the following reaction: 4NO_(2)(g) + O_(2)(g) to 2N_(2)O_(5)(g),De

1.	In the following reaction: 4NO_(2)(g) + O_(2)(g) to 2N_(2)O_(5)(g),DeltaH = - 110 kJ, if N_(2)o_(5)(s) is formed instead of N_(2)O_(5)(g) in the reaction, the enthalpy change (in kJ) would be (enthalpy of sublimation of N_(2)O_(5)(s) is + 53 kJ mol^(-1)).
Answer» `-216` `-162` `+108` `+216` Solution :(i) `4NO_(2)(g) + O_(2)(g) to 2N_(2)O_(5)(g), DeltaH = -110kJ` GIVEN that (ii) `N_(2)O_(5)(s) to N_(2)O_(5)(g) , DeltaH = 53 kJ"mol"^(-1)` (iii)`2N_(2)O_(5)(s) to 2N_(2)O_(5)(g), DeltaH = 106 kJ"mol"^(-1)` Subtracting eq.(iii) from (i), `4NO_(2)(g) +O_(2)(g) to 2N_(2)O_(5)(s)` then `DeltaH = -110-106 = -216 kJ"mol"^(-1)`