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Enthalpies of Formation Find the Value of Triangle Rh for the Reaction - Chemistry

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

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

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Solution 1

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

`triangle_r H =  sum triangle_f H("products") - sum triangle_f H("reactants")`

For the given reaction,

N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)

`triangle_rH = [{triangle_f H(N_2O) + 3triangle_f H(CO_2)}-{triangle_fH(N_2O_4) + 3triangle_fH(CO)}]`

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

`triangle_rH = [{81 kJ mol^(-1) + 3(-393)kJ mol^(-1)}-{9.7 kJ mol^(-1) + 3(-110)kJ mol^(-1)}]`

`triangle_rH = -777.7 kJ mol^(-1)`

Hence, the value of Δrfor the reaction is `- 777.7 kJ mol^(-1)`

Solution 2

Enthalpy of reaction (∆r,H) = [81 + 3 (- 393)] – [9.7 + 3 (- 110)]

= [81 – 1179] – [9.7 – 330] = – 778 kj mol-1

Concept: Enthalpy Change, ∆_rH of a Reaction - Reaction Enthalpy - Standard Enthalpy of Reactions
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 6 Thermodynamics
Q 12 | Page 183
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