Question

Consider the reaction
$4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g),$
$\Delta_rH$ = – 111 kJ.
If N2O5 (s) is formed instead of N₂O₅ (g) in the above reaction, the $\Delta_rH$ value will be :(given, $\Delta H$ of sublimation for N₂O₅ is –54 kJ mol⁻¹)
(a) + 54 kJ
(b) + 219 kJ
(c) – 219 J
(d) – 165 kJ

Answer 1

-219kj Becaz N2O5 (s) ko sublimation karne 54 kj /mol sublimation energy N2O5(s) to N2O5 (g) ke liye +54 then for reverse it is -54 for 1 molecule and 2×54= -108 then -111+(-108)= -

219