Consider the reaction: 4 NO 2( g )+ O 2( g ) arrow 2 N 2 O 5( g ) Δr H =-111 kJ If N 2 O 5( s ) is formed instead of N 2 O 5( g ) in the above reaction, the Δr H value will be: (given, Δ H of sublimation for N 2 O 5 is 54 kJ mol -1 )

Question

Consider the reaction: 4 NO 2( g )+ O 2( g ) arrow 2 N 2 O 5( g ) Δr H =-111 kJ If N 2 O 5( s ) is formed instead of N 2 O 5( g ) in the above reaction, the Δr H value will be: (given, Δ H of sublimation for N 2 O 5 is 54 kJ mol -1 ) (A) + 54 kJ (B) + 219 kJ (C) - 219 kJ (D) - 165 kJ

Tardigrade Admin · Accepted Answer

- 111 - 54 = Δ H' Δ H' =- 165 KJ

Q. Consider the reaction :
$4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g)$
$Δ_{r} H = - 111 k J$
If $N_{2} O_{5} (s)$ is formed instead of $N_{2} O_{5} (g)$ in the above reaction, the $Δ_{r} H$ value will be :
(given, $Δ H$ of sublimation for $N_{2} O_{5}$ is $54 k J m o l^{- 1}$ )

$+ 54 k J$

$+ 219 k J$

$- 219 k J$

$- 165 k J$

$- 111 - 54 = Δ H^{'}$
$Δ H^{'} = - 165 K J$

Q. Consider the reaction : 4NO2​(g)+O2​(g)→2N2​O5​(g) Δr​H=−111kJ If N2​O5​(s) is formed instead of N2​O5​(g) in the above reaction, the Δr​H value will be : (given, ΔH of sublimation for N2​O5​ is 54kJmol−1 )

+54kJ

+219kJ

−219kJ

−165kJ