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12N2(g)+O2(g)→NO2(g); ΔrH∘=−40kJ/mol Given: Cp,mNO2,g=40J/mol/K;CpmO2,g=30JK−1mol−1Cp,mN2(g)=30JK−1mol−1What is the enthalpy of formation of NO2 (g) at 1298 K?

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a

- 40 kJ/mol

b

-50 kJ/mol

c

- 45 kJ/mol

d

- 6 kJ/mol

answer is C.

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Detailed Solution

ΔrHT=ΔrH∘+∫2981298 ΔrCPdT At 1298K,ΔrH=−40kJ−5ΔT=−40kJ−5×1000×10−3kJ=−45kJ/mol

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