Given,NO(g) +O3 (g) → NO2(g) +O2 (g) Δ H1 =-198.9 KJ/mol O3( g)arrow 3 2 O2(g); Δ H =-142.3 KJ mol O2(g)arrow O2 (g); Δ H3 =+495.0 KJ mol The enthalpy change (Δ H) for the following reaction is NO(g)+O(g) arrow NO2(g)

Question

Given,NO(g) +O3 (g) → NO2(g) +O2 (g) Δ H1 =-198.9 KJ/mol O3( g)arrow 3 2 O2(g); Δ H =-142.3 KJ mol O2(g)arrow O2 (g); Δ H3 =+495.0 KJ mol The enthalpy change (Δ H) for the following reaction is NO(g)+O(g) arrow NO2(g) (A) -3041 kJ/mol (B) +304.1 kJ/mol (C) -4031 kJ/mol (D) +403.1 kJ/mol

Tardigrade Admin · Accepted Answer

Given NO(g) +O3 (g) → NO2(g) +O2 (g) Δ H1 =-198.9 KJ/mol O3 arrow (3/2) O2(g) Δ H2 =-142.3 KJ/ mol O2arrow (3/2) O2 (g) Δ H3 +=495.0 KJ /mol For the reaction NO(g)+O(g) arrow NO2(g) Δ H=Δ H1 -Δ H2 -(1/2) Δ H3 =-198.9-(-142.3)- (1/2)×495 =-304.1 kJ/ mol

Q. Given,NO(g)+O3​(g)→NO2​(g)+O2​(g) ΔH1​=−198.9KJ/mol O3​(g)→32O2​(g);ΔH=−142.3KJ mol O2​(g)→O2(g);ΔH3​=+495.0KJmol The enthalpy change (ΔH) for the following reaction is NO(g)+O(g)→NO2​(g)

-3041 kJ/mol

+304.1 kJ/mol

-4031 kJ/mol

+403.1 kJ/mol

Given NO(g)+O3​(g)→NO2​(g)+O2​(g) ΔH1​=−198.9KJ/mol O3​→23​O2​(g)ΔH2​ =−142.3KJ/ mol O2​→23​O2​(g)ΔH3​+=495.0KJ/mol For the reaction NO(g)+O(g)→NO2​(g) ΔH=ΔH1​−ΔH2​−21​ΔH3​ =−198.9−(−142.3)−21​×495 =−304.1kJ/ mol

Q. Given, $NO (g) + O_{3} (g) \to N O_{2} (g) + O_{2} (g)$
$Δ H_{1} = - 198.9 K J /$ mol
$O_{3} (g) \to 32 O_{2} (g); Δ H = - 142.3 K J$ mol
$O_{2} (g) \to O 2 (g); Δ H_{3} = + 495.0 K J$ mol
The enthalpy change $(Δ H)$ for the following reaction is
$NO (g) + O (g) \to N O_{2} (g)$