Q. 124.2( 39 Votes )

Enthalpies of for

Answer :

Given: For the given reaction:

N2O4(g) + 3CO(g) N2O(g) + 3CO2(g)

Enthalpy of formation (ΔHf (CO)) = -110 kJ mol-1

ΔHf (CO2) = -393 kJ mol-1

ΔHf (N2O) = 81kJ mol-1

ΔHf (N2O4) = 9.7 kJ mol–1

Now, to calculate ΔrH for the reaction, we apply the formula given below:

ΔrH = (Sum of enthalpy of formation of products) – (sum of enthalpy of reactants)

or, ΔrH = ∑ ΔfH (Products) - ∑ ΔfH (Reactants)

In the given reaction,

N2O4(g) + 3CO(g) N2O(g) + 3CO2(g)

Products are N2O and CO2

Reactants are N2O4 and CO

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

By substituting the values given, we get

ΔrH = [81 kJ/mol + 3 × (-393 kJ/mol)] – [9.7 kJ/mol + 3 × (-110 kJ/mol)]

ΔrH = -777.7 kJ/mol

Thus, the value of ΔrH for the reaction is -777.7 kJ/mol.

Note: Enthalpy of reaction (ΔrH) is actually the difference between the enthalpies of the products and the reactants when the quantities of the reactants indicated by the chemical equation have completely reacted.

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