Physical Chemistry Plus Mastering Chemistry With Etext -- Access Card Package (3rd Edition) (engel Physical Chemistry Series)
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Chapter 4, Problem 4.7NP

(a)

Interpretation Introduction

Interpretation:

The bond enthalpy and energy should be calculated for the CH bond in CH4 .

Concept Introduction:

The mathematical expression for the standard enthalpy change value at room temperature is:

  ΔH°R=nH°(products)pH°(reactants)

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The mathematical expression for the standard enthalpy is:

According Hess’s law,

  ΔHo=ΔH1o+ΔH2o+ΔH3o

The mathematical expression for standard internal energy change is:

  ΔHo=ΔUo+ΔPV

  ΔUo=ΔHoΔnRT

(b)

Interpretation Introduction

Interpretation:

The bond enthalpy and energy should be calculated for the CC bond in C2H6 .

Concept Introduction:

The mathematical expression for the standard enthalpy change value at room temperature is:

  ΔH°R=nH°(products)pH°(reactants)

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The mathematical expression for the standard enthalpy is:

According Hess’s law,

  ΔHo=ΔH1o+ΔH2o+ΔH3o

The mathematical expression for standard internal energy change is:

  ΔHo=ΔUo+ΔPV

  ΔUo=ΔHoΔnRT

(c)

Interpretation Introduction

Interpretation:

The bond enthalpy and energy should be calculated for the C=C bond in C2H4 .

Concept Introduction:

The mathematical expression for the standard enthalpy change value at room temperature is:

  ΔH°R=nH°(products)pH°(reactants)

Where, n and p represents the coefficients of reactants and products in the balanced chemical equation.

The mathematical expression for the standard enthalpy is:

According Hess’s law,

  ΔHo=ΔH1o+ΔH2o+ΔH3o

The mathematical expression for standard internal energy change is:

  ΔHo=ΔUo+ΔPV

  ΔUo=ΔHoΔnRT

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For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3) → Give the expression for the acceptable rate. → → (A). d[N205] dt == 2k,k₂[N₂O₂] k₁+k₁₂ (B). d[N2O5] =-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³ dt (C). d[N2O5] =-k₁[N₂O] + k [NO] - k₂[NO] [NO] d[N2O5] (D). = dt = -k₁[N2O5] - k¸[NO][N₂05] dt Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.
For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3) → Give the expression for the acceptable rate. → → (A). d[N205] dt == 2k,k₂[N₂O₂] k₁+k₁₂ (B). d[N2O5] =-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³ dt (C). d[N2O5] =-k₁[N₂O] + k [NO] - k₂[NO] [NO] d[N2O5] (D). = dt = -k₁[N2O5] - k¸[NO][N₂05] dt Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.
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