Chapter 9, Problem 9.94QA

Interpretation Introduction

To calculate:

The enthalpy changes of the following reactions under standard conditions using average bond energies.

a) $C{O}_2\left(g\right)+H_2\left(g\right)\to H_{2}O\left(g\right)+CO\left(g\right)$

b) $N_2\left(g\right)+O_2\left(g\right)\to 2 NO\left(g\right)$

c) $C\left(s\right)+C{O}_2\left(g\right)\to 2CO\left(g\right)$

The heat of sublimation of graphite $C\left(s\right)$ is $719 kJ/mol$.

Answer to Problem 9.94QA

Solution:

a) $C{O}_2\left(g\right)+H_2\left(g\right)\to H_{2}O\left(g\right)+CO\left(g\right)$; ${∆H}_{rxn}^o=36 kJ$

b) $N_2\left(g\right)+O_2\left(g\right)\to 2 NO\left(g\right)$; ${∆H}_{rxn}^o=87 kJ$

c) $C\left(s\right)+C{O}_2\left(g\right)\to 2CO\left(g\right)$; ${∆H}_{rxn}^o=173 kJ$

Explanation of Solution

1) Concept:

Bond energy is the amount of energy required when one mole of bonds in the gas phase is broken. This can be used to calculate the heat of reaction (enthalpy change) by combining the $∆H$ values (bond energies) for bonds that are broken and bonds that are formed.

$∆H$ values are positive for bonds broken.

$∆H$ values are negative for bonds formed.

${∆H}_{rxn}^o=\sum {∆H}_{f bond breaking}^o-\sum {∆H}_{f bond forming}^o$

2) Formula:

${∆H}_{rxn}^o=\sum {∆H}_{f bond breaking}^o-\sum {∆H}_{f bond forming}^o$

3) Given:

The heat of sublimation of graphite $C\left(s\right)$ is $719 kJ/mol$.

Bond energies from Appendix 4: Table A4.1

Bonds	Bond energy ($kJ/mol$)
$C=O$	$743$
$H-H$	$436$
$O-H$	$463$
$C\equiv O$	$1072$
$N\equiv N$	$945$
$O=O$	$498$
$N\equiv O$	$678$

4) Calculation:

a) $C{O}_2\left(g\right)+H_2\left(g\right)\to H_{2}O\left(g\right)+CO\left(g\right)$

	Bonds	Number of bonds (mol)	Bond energy (kJ/mol)	$\Delta H$
Bonds broken	$C=O$	2	$799 kJ/mol$	$2 mol \times 799 kJ/mol$
	$H-H$	1	$436 kJ/mol$	$1 mol \times 436 kJ/mol$
Bonds formed	$O-H$	2	$463 kJ/mol$	$-\left(2 mol \times 463 kJ/mol\right)$
	$C\equiv O$	1	$1072 kJ/mol$	$-\left(1 mol \times 1072 kJ/mol\right)$

${∆H}_{rxn}^o=\left[\left(2 mol \times 799 kJ/mol\right)+\left(1 mol \times 436 kJ/mol\right)\right]-\left[\left(2 mol \times 463 kJ/mol\right)+\left(1 mol \times 1072 kJ/mol\right)\right]$

${∆H}_{rxn}^o=2034 kJ-1998 kJ=36 kJ$

b) $N_2\left(g\right)+O_2\left(g\right)\to 2 NO\left(g\right)$

	Bonds	Number of bonds (mol)	Bond energy (kJ/mol)	$\Delta H$
Bonds broken	$N\equiv N$	1	$945 kJ/mol$	$1 mol \times 945 kJ/mol$
	$O=O$	1	$498 kJ/mol$	$1 mol \times 498 kJ/mol$
Bonds formed	$N\equiv O$	2	$678 kJ/mol$	$-\left(2 mol \times 678 kJ/mol\right)$

${∆H}_{rxn}^o=\left[\left(1 mol \times 945 kJ/mol\right)+\left(1 mol \times 498 kJ/mol\right)\right]-\left[\left(2 mol \times 678 kJ/mol\right)\right]$

${∆H}_{rxn}^o= 661 kJ-1356 kJ=87 kJ$

c) $C\left(s\right)+C{O}_2\left(g\right)\to 2CO\left(g\right)$

$C\left(s\right)\to C\left(g\right)$; ${∆H}_{sub}=719 kJ/mol$

	Bonds	Number of bonds (mol)	Bond energy (kJ/mol)	$\Delta H$
Bonds broken	$C=O$	2	$799 kJ/mol$	$2 mol \times 799 kJ/mol$
Bonds formed	$C\equiv O$	2	$1072 kJ/mol$	$-\left(2 mol \times 1072 kJ/mol\right)$

Here, solid carbon is converted to gaseous form. Therefore, the heat of sublimation also need to be taken into consideration.

${∆H}_{rxn}^o=\left[\left(1 mol \times 719 kJ/mol\right)+\left(2 mol \times 799 kJ/mol\right)\right]-\left[2 mol \times 1072 kJ/mol\right]$

${∆H}_{rxn}^o= 2317 kJ-2144 kJ=173 kJ$

Conclusion:

The enthalpy change is the difference between the $∆H$ values (bond energies) for bonds that are broken and bonds that are formed.