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Question: \(\frac{1}{2}\)N<sub>2</sub> (g) + O<sub>2</sub> (g) ¾® NO<sub>2</sub> (g) D<sub>r</sub>Hŗ = – 40 kJ...

12\frac{1}{2}N2 (g) + O2 (g) ¾® NO2 (g) DrHŗ = – 40 kJ/mol

Given :Cp, m (NO2, g) = 40 J/mol/K ;

Cp, m (O2, g) = 30 JK–1 mol–1

Cp, m N2 (g) = 30 JK–1mol–1

What is the enthalpy of formation of NO2 (g) at 1298 K ?

A

– 40 kJ/mol

B

–50 kJ/mol

C

– 45 kJ/mol

D

– 6 kJ/mol

Answer

– 45 kJ/mol

Explanation

Solution

Dr HT = DrH0 + ∫2981298ΔrCPdT\int_{298}^{1298}{\Delta_{r}C_{P}dT}At 1298 K DrH = – 40 kJ – 5

DT = – 40 kJ – 5 × 1000 × 10–3 kJ

= – 45 kJ/mol