Chapter 19, Problem 19.112QA

Interpretation Introduction

To calculate:

The fuel values of liquid formamide and methyl formate.

Answer to Problem 19.112QA

Solution:

Fuel value of liquid formamide = $12.00 kJ/g.$

Fuel value of liquid methyl formate = $16.11 kJ/g.$

Explanation of Solution

1) Concept:

We are asked to calculate the fuel value for liquid formamide and methyl formate. The enthalpy of formation for both in the units $kJ/mol$ is given. Fuel value is the amount of heat given off when one gram of fuel is burned. We are given enthalpy of formation for formaldehyde and formic acid; we will write the combustion reactions of formaldehyde and formic acid and using the $∆H_f^0$ values for reactants and products, we calculate the $∆H{°}_{comb}$ values for both. Fuel value for compounds are measured in $kJ/g$. So, we would convert the energies from $kJ/mol$ to $kJ/g$ using molar masses for formaldehyde and formic acid to get the fuel values for both.

2) Given:

Standard enthalpy of formation for liquid formamide, $∆H{°}_f=-251 kJ/mol$

Standard enthalpy of formation for liquid methyl formate $∆H{°}_f=-391 kJ/mol$

Mass of one mole of formamide = $45.0407g$

Mass of one mole of methyl formate = $60.0516 g$

3) Calculations:

Combustion reaction of formamide can be written as:

$4C{H}_{3}NO\left(l\right)+9O_2\left(g\right)\to 4C{O}_2\left(g\right)+6H_{2}O\left(l\right)+4N{O}_2\left(g\right)$

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

From Appendix 4, the $∆H^0$ values for $C{O}_2\left(g\right)$,$H_{2}O\left(l\right)$ and $N{O}_2\left(g\right)$ are $-393.5 kJ/mol$,$-285.8 kJ/mol$ and $33.2 kJ/mol$.

Using these values and the above equation, let’s calculate the $∆H_{comb}^0$ value for formamide as:

$∆H{°}_{comb}=\left[\left(6mol\times \frac{-285.8 kJ}{mol}\right)+ \left(4mol\times \frac{-393.5 kJ}{mol}\right)+\left(4mol\times \frac{33.2 kJ}{mol}\right)\right]-\left[\left(4mol \times \frac{-251 kJ}{mol}\right)+\left(9mol\times \frac{0 kJ}{mol}\right)\right]$

$∆H{°}_{comb}=\left(-3156 kJ\right)-\left(-1004kJ\right)= -2152 kJ$

Calculating fuel value for formamide as:

$\frac{-2152 kJ}{4 mol}\times \frac{1mol}{45.0407 g}=11.94 kJ/g$

Thus, the fuel value of formamide is $12.0\frac{kJ}{g}$

Combustion reaction of methyl formate is written as:

${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{O}}_2\left(l\right)+2O_2\left(g\right)\to 2C{O}_2\left(g\right)+{2H}_{2}O\left(l\right)$

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

From Appendix 4, the $∆H^0$ values for $C{O}_2\left(g\right)$ and $H_{2}O\left(l\right)$ are $-393.5 kJ/mol$ and $-285.8 kJ/mol$ respectively.

Using these values and the above equation, let’s calculate the $∆H_{comb}^0$ value for methyl formate as:

$∆H{°}_{comb}=\left[\left(2mol\times \frac{-285.8kJ}{mol}\right)+ \left(2mol\times \frac{-393.5 kJ}{mol}\right)\right]-\left[\left(1 mol \times \frac{-391 kJ}{mol}\right)+\left(2mol\times \frac{0kJ}{mol}\right)\right]$

$∆H{°}_{comb}=\left(-1358.6 kJ\right)-\left(-391.0\right)= -967.6 kJ$

Calculating fuel value for methyl formate as:

$\frac{967.6 kJ}{1 mol}\times \frac{1mol}{60.0516 g}= 16.11 kJ/g$

Thus, the fuel value of methyl formate is $16.11\frac{kJ}{g}$

Conclusion:

Fuel value of liquid formamide is smaller than that of liquid methyl formate.