Chapter 9, Problem 9.91QA

Interpretation Introduction

To find:

The enthalpy changes ($∆H_{rxn}^0$) of given reactions under standard conditions.

Answer to Problem 9.91QA

Solution:

a) $-93 kJ$

b) $90 kJ$

c) $338 kJ$

Explanation of Solution

1) Concept:

For chemical reactions occurring in gas phase, the impact of intermolecular forces on the energies of reactants and products can be ignored. For gas phase reactions, the bond energies are usually very close to their average values. So the values of average bond energies can be used to calculate $∆H_{rxn},$ which is quite reliable.

2) Formula:

${∆H}_{rxn}=Ʃ{∆H}_{\mathit{b}\mathit{o}\mathit{n}\mathit{d} breaking}-Ʃ∆H_{bond making}$

Positive values of $∆H$ are taken in case of bond breaking, and negative values of $∆H$ are considered for bond forming.

3) Given:

The required average bond energy values are taken from Table A4.1 of Appendix 4.

Bond	Bond energy (kJ/mol)
	$945$
H-H	$436$
N-H	$391$
N-N	$163$
	$498$
N=N	$418$
N=O	$678$

4) Calculations:

a) The given balanced reaction is

$N_2\left(g\right)+{3H}_2\left(g\right)\to {2NH}_3\left(g\right)$

The reaction involves the formation of two moles of ammonia from its constituent elements in their standard states.

First, we have to draw the Lewis structures of reactants and products into the balanced chemical equation to get the number of bonds involved in each reactant and product.

+   ⤍

 We can construct a table using the appropriate bond energies:

	Bonds	Number of bonds(mol)	Bond energy kJ/mol	$∆H$
Bonds broken		$1$	$945$	$1mol\times \frac{945kJ}{mol}$
Bonds broken	$H-H$	$3$	$436$	$3mol\frac{436kJ}{mol}$
Bonds formed	$N-H$	$6$	$391$	$-6mol\times \frac{391kJ}{mol}$

We can get the value of $∆H_{rxn}$ by summing the values in the right-hand column.

${∆H}_{rxn}=\left(1mol\times \frac{945kJ}{mol}\right)+\left(3mol\times \frac{436kJ}{mol}\right)-\left(6 mol \times \frac{391 kJ}{mol}\right)-93kJ$

b) The given balanced reaction is

$N_2\left(g\right)+{2H}_2\left(g\right)⤍H_{2}NN{H}_2\left(g\right)$

The reaction involves the formation of one mole of hydrazine from its constituent elements in their standard states.

First, we have to draw the Lewis structures of reactants and products into the balanced chemical equation to get the number of bonds involved in each reactant and product.

+  ⤍

 We can construct a table using the appropriate bond energies:

	Bonds	Number of bonds(mol)	Bond energy (kJ/mol)	$∆H$
Bonds broken		$1$	$945$	$1mol\times \frac{945kJ}{mol}$
Bonds broken	H-H	$2$	$436$	$2mol\times \frac{436kJ}{mol}$
Bonds formed	N-N	$1$	$163$	$-1mol\times \frac{163kJ}{mol}$
Bonds formed	N-H	$4$	$391$	$-4mol\times \frac{391kJ}{mol}$

We can get the value of $∆H_{rxn}$ by summing the values in the right-hand column.

${∆H}_{rxn}=\left[\left(1mol\times \frac{945kJ}{mol}\right)+\left(2mol\times \frac{436kJ}{mol}\right)\right]-\left[\left(1mol\times \frac{163kJ}{mol}\right)+\left(4mol\times \frac{391kJ}{mol}\right)\right]$

$=90kJ$

c)The given balanced reaction is:

${2N}_2\left(g\right)+O_2\left(g\right)\to {2N}_{2}O\left(g\right)$

The reaction involves the formation of two moles of nitrous oxide gas from its constituent elements in their standard states.

First, we have to draw the Lewis structures of reactants and products into the balanced chemical equation to get the number of bonds involved in each reactant and product.

+ ⤍

We can construct a table using the appropriate bond energies:

	Bonds	Number of bonds(mol)	Bond energy (kJ/mol)	∆H
Bonds broken		2	945	$2mol\times \frac{945kJ}{mol}$
Bonds broken		1	498	$1mol\times \frac{498kJ}{mol}$
Bonds formed	N=N	2	418	$-2mol\times \frac{418kJ}{mol}$
Bonds formed	N=O	2	607	$-2mol\times \frac{607kJ}{mol}$

We can get the value of $∆H_{rxn}$ by summing the values in the right-hand column.

${∆H}_{rxn}=\left[\left(2mol\times \frac{945KJ}{mol}\right)+\left(1mol\times \frac{498KJ}{mol}\right)\right]-\left[\left(2mol\times \frac{418KJ}{mol}\right)+\left(2mol\times \frac{607KJ}{mol}\right)\right]$

${∆H}_{rxn}=338 kJ$

Conclusion:

The standard enthalpy change of a reaction can be calculated using the standard average bond energies per mole for each reactant and product.