Question
Asked Dec 3, 2019
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Determine the standard enthalpy of formation for N2O given the following information about the formation of NO2 under standard conditions, and ΔHf (NO2) = +33.2 kJ/mol.

2N2O(g)+3O2(g) ---> 4NO2(g)             ΔHrxn= -31.4 kJ

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Expert Answer

Step 1

Hess's law

The enthalpy change of a reaction is calculated by subtraction of sum of standard enthalpy of formation of reactants from sum of standard enthalpy of formation of products.

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.- Σ ΔΗρπόκ)Σ ΔΗ, ΔΗ, LEφro oucs) reaction. f(reactants)

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Step 2

At the standard state, the standard enthalpy of formation all substances are zero.

Given:

ΔHf (NO2) = +33.2 kJ/mol, ΔHrxn= -31.4 kJ.

The balanced reaction and enthalpy values are,

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ΑΝΟ4 2N,Ο Τ5 Ο53) 4ΝO, ΔΗxiαΣ ΔΗ ; ανεω ΣΔΗ (mm ) τού μct. =-31.4kJ Τex = 33.2 kJ ΔΗ. ΔΗo. =0 kJ

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Step 3

Substitute the given values in above equation to get st...

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ΔΗ-ΔΗΝΟ, -ΔΗ -ΔΗ ΔΗ -ΔΝΟ -ΝO, rex 2(ΔΗ0 = -312kI-4(3 3.2 k) -101.6 -50.8 kJ 2 - ΔΗ N,Ο ΔΗΝ.-50.8 kJ

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Chemistry

Chemical Thermodynamics

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