Chapter 19, Problem 19.113QA

Interpretation Introduction

To calculate:

 $∆H{°}_{rxn}$ for the given reactions of methanogenic bacteria.

Answer to Problem 19.113QA

Solution:

a) $∆H{°}_{rxn}=16.2 kJ$

b) $∆H{°}_{rxn}=-126.9 kJ$

Explanation of Solution

1) Concept:

We are given two balanced reactions of methanogenic bacteria and asked to find the $∆H{°}_{rxn}$ for each given reaction. We can find the standard enthalpy formation values for $C{H}_4\left(g\right), C{O}_2\left(g\right), C{H}_{3}COOH \left(l\right)and H_{2}O\left(l\right)\_ can find out the standard enthalpy formation values for$ from the Appendix and find the standard enthalpy for the reaction using the following equation:

$∆H{°}_{rxn}= \sum n_{products}∆H_{f, products}^0- \sum n_{reactants}∆H_{f, reactants}^0$

2) Formula:

$∆H{°}_{rxn}= \sum n_{products}∆H_{f, products}^0- \sum n_{reactants}∆H_{f, reactants}^0$

3) Given:

i) Standard enthalpy formation value for formic acid $∆H{°}_f=-425.0 kJ/mol$

ii) Standard enthalpy formation value for $C{H}_4\left(g\right)= -74.8 kJ/mol$(from Appendix 4)

iii) Standard enthalpy formation value for $C{O}_2\left(g\right)= -393.5 kJ/mol$(from Appendix 4)

iv) Standard enthalpy formation value for $H_{2}O\left(l\right)= -285.8 kJ/mol$(from Appendix 4)

v) Standard enthalpy formation value for $C{H}_{3}COOH\left(l\right)= -484.5 kJ/mol$(from Appendix 4)

vi) $C{H}_{3}COOH \left(l\right)\to C{H}_4\left(g\right)+C{O}_2\left(g\right)$

vii) $4HCOOH\left(l\right)\to C{H}_4\left(g\right)+3C{O}_2\left(g\right)+2H_{2}O\left(l\right)$

4) Calculations:

(1) Calculating standard enthalpy for the first given reaction:

$C{H}_{3}COOH \left(l\right)\to C{H}_4\left(g\right)+C{O}_2\left(g\right)$

Calculating $∆H{°}_{rxn}$ for this reaction using the formula:

$∆H{°}_{rxn}= \sum n_{products}∆H_{f, products}^0- \sum n_{reactants}∆H_{f, reactants}^0$

Plugging in the values in the equation, we get:

$∆H{°}_{rxn}=\left[\left(1mol C{H}_4\times \frac{-74.8kJ}{mol C{H}_4}\right)+ \left(1mol C{O}_2\times \frac{-393.5kJ}{mol C{O}_2}\right)\right]-\left[\left(1 mol C{H}_{3}COOH\times \frac{-484.5kJ}{mol C{H}_{3}COOH}\right)\right]$

$∆H{°}_{rxn}=\left(-468.3 kJ\right)-\left(-484.5 kJ\right)$

$∆H{°}_{rxn}= -468.3 kJ+484.5 kJ=16.2kJ$

2) Calculating standard enthalpy for second reaction:

$4HCOOH\left(l\right)\to C{H}_4\left(g\right)+3C{O}_2\left(g\right)+2H_{2}O\left(l\right)$

$∆H{°}_{rxn}= \sum n_{products}∆H_{f, products}^0- \sum n_{reactants}∆H_{f, reactants}^0$

Plugging in the values in the equation, we get:

$∆H{°}_{rxn}=\left[\left(2mol{H}_{2}O\times \frac{-285.8kJ}{mol{H}_{2}O}\right)+ \left(3mol C{O}_2\times \frac{-393.5kJ}{mol C{O}_2}\right)+(1mol C{H}_4 \times \frac{-74.8kJ}{mol C{H}_4}\right]-\left[\left(4 mol HCOOH \times \frac{-425.0kJ}{mol HCOOH}\right)\right]$

$∆H{°}_{rxn}=\left(-571.6 kJ -1180.5 kJ-74.8 kJ\right)-\left(-1700 kJ \right)$

$∆H{°}_{rxn}= -1826.9 kJ+1700kJ=-126.9 kJ$

Conclusion:

Using standard enthalpy values from Appendix 4, we calculated the enthalpy of reactions of methanogenic bacteria.