Question

Consider the reaction :

4NO2(g) + O2(g) $\rightarrow$2N2O5(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.
  • 54kJ
• B.
  • 219 kJ
• C. – 219 kJ
• D. – 165 kJ

Answer 1

Answer: (D)

– 165 kJ

Answer 2

– 111 – 54 = $\Delta$H'

$\Delta$H' = – 165 KJ