SELECT * FROM question_mgmt as q WHERE id=2798 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=47 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ΔrH for the reaction:

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

Answer:

ΔrH for a reaction is defined as the difference between ΔfH value of products and ΔfH value of reactants.

ΔrH = \sum  ΔfH (products) - \sum  ΔfH (reactants)

For the given reaction,

N2O4(g)+ 3CO(g)  →  N2O(g) + 3CO2(g)

ΔrH = [{ΔfH (N2O) + 3ΔfH(CO2)}  -  {ΔfH(N2O4) + 3ΔfH(CO)}]

Substituting the values of ΔfH for N2O, CO2, N2O4,and CO from the question, we get:

ΔrH = [{81 KJ mol-1 + 3(-393) KJ mol-1} - {9.7KJ mol-1 + 3(-110)KJ mol-1}]

ΔrH = -777.7 KJ mol-1

Hence, the value of ΔrH for the reaction is -777.7 KJ mol-1.


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