Consider the reaction, 4NO2(g)+O2(g)→2N2O5(g),ΔrH=−111kJ. If N2O5(s) is formed instead of N2O5(g) in the above reaction, the ΔrH value will be ( Given, ΔH of sublimation for N2O5 is 54kJmol−1 )
A
−165kJ
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B
+54kJ
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C
+219kJ
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D
−219kJ
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Solution
The correct option is A−219kJ As 2moles of N2O5 is formed net sublimation energy =108kJ. NO2+O2(g)⟶2N2O5(s)ΔHr=? Adding 2N2O5(s)Δ⟶2N2O5(g)ΔH=+108kJ 4NO2+O2⟶2N2O5(g)ΔHnet=−111kJ ⟹ΔHr+ΔH=ΔHnet ⟹ΔHr=−111−108 ⟹ΔHr=−219kJ