Chapter 19, Problem 19.76QA

Interpretation Introduction

To determine:

The $∆H_{rxn}^0$ value of the following combustion reactions of methylamine:

a) $4 C{H}_{3}N{H}_2 \left(g\right)+13 O_2\left(g\right) \to 4 C{O}_2\left(g\right)+4 N{O}_2\left(g\right)+10 H_{2}O \left(l\right)$

b) $4 C{H}_{3}N{H}_2 \left(g\right)+6 O_2\left(g\right) \to 4 C{O}_2\left(g\right)+4 N{H}_3\left(g\right)+4 H_{2}O \left(l\right)$

Answer to Problem 19.76QA

Solution:

a)  For the reaction $4 C{H}_{3}N{H}_2 \left(g\right)+13 O_2\left(g\right) \to 4 C{O}_2\left(g\right)+4 N{O}_2\left(g\right)+10 H_{2}O \left(l\right)$, the $∆H_{rxn}^0=-4207.2 kJ$.

b) For the reaction $4 C{H}_{3}N{H}_2 \left(g\right)+6 O_2\left(g\right) \to 4 C{O}_2\left(g\right)+4 N{H}_3\left(g\right)+4 H_{2}O \left(l\right),$ the $∆H_{rxn}^0=-2809.6 kJ$.

Explanation of Solution

1) Concept:

To calculate the standard enthalpy of the reaction, we need to use the heat of formation of reactant and product. We will calculate the $∆H_{rxn}^0$ from the thermodynamic value, $∆H_f^0,$ of the compounds present in the reaction.

2) Formula:

$∆H_{rxn}^0 = \sum m \times ∆H_{f, product}^0-\sum n \times ∆H_{f, reactant}^0$

3) Given:

i) ${CH}_{3}N{H}_2 \left(g\right)$,   $∆H_f^0 =-23.0 kJ/mol$$$$$$$

ii) ${CO}_2 \left(g\right)$,   $∆H_f^0 = -393.5 kJ/mol$$$$$$$

iii) $H_{2}O \left(l\right)$,   $∆H_f^0 = -285.8 kJ/mol$$$$$$$

iv) $O_2 \left(g\right)$,   $∆H_f^0 = -0.0 kJ/mol$$$$$$$

v) $N{H}_3 \left(g\right)$,   $∆H_f^0 = -46.1 kJ/mol$$$$$$$

vi) ${NO}_2 \left(g\right)$,   $∆H_f^0 = 33.2 kJ/mol$$$$$$$

4) Calculations:

a) The combustion reaction of methylamine is

$4 {CH}_{3}N{H}_2 \left(g\right)+ 13 O_2\left(g\right)\to 4 C{O}_2\left(g\right)+4 N{O}_2\left(g\right)+10 H_{2}O \left(l\right)$

Calculate the $∆H_{rxn}^0$ of the above combustion reaction by using the thermodynamic value given from Appendix 4.

$∆H_{rxn}^0 = \sum m \times ∆H_{f, product}^0-\sum n \times ∆H_{f, reactant}^0$

$∆H_{rxn}^0 = \left[4 \times \left(-393.5 \frac{kJ}{mol}\right)+ 4\times \left(33.2 \frac{kJ}{mol}\right)+10 \times \left(-285.8 \frac{kJ}{mol}\right)\right]-\left[4 \times \left(-23.0 \frac{kJ}{mol}\right)+13 \times \left(0.0 \frac{kJ}{mol}\right)\right]$

$∆H_{rxn}^0 =\left(-4299.2\right) kJ-\left(-92.0\right) kJ= -4207.2 kJ$

The heat of the reaction $∆H_{rxn}^0$ is $-4207.2 kJ$

b) The combustion reaction of methylamine is

$4 {CH}_{3}N{H}_2 \left(g\right)+ 6 O_2\left(g\right)\to 4 C{O}_2\left(g\right)+4 N{H}_3\left(g\right)+4 H_{2}O \left(l\right)$

Calculate the $∆H_{rxn}^0$ of the above combustion reaction by using the thermodynamic value given from Appendix 4.

$∆H_{rxn}^0 = \sum m \times ∆H_{f, product}^0-\sum n \times ∆H_{f, reactant}^0$

$∆H_{rxn}^0 = \left[4 \times \left(-393.5 \frac{kJ}{mol}\right)+ 4\times \left(-46.1 \frac{kJ}{mol}\right)+4 \times \left(-285.8 \frac{kJ}{mol}\right)\right]-\left[4 \times \left(-23.0 \frac{kJ}{mol}\right)+6 \times \left(0.0 \frac{kJ}{mol}\right)\right]$

$∆H_{rxn}^0 =\left(-2901.6\right) kJ-\left(-92.0\right) kJ= -2809.6 kJ$

The heat of the reaction $∆H_{rxn}^0$ is $-2809.6 kJ$

Conclusion:

The heat of reaction is calculated from the thermodynamic data and combustion reaction.