Chapter 9, Problem 9.111QA

Interpretation Introduction

To find:

a) What is the standard heat of combustion of acetylene, where ${∆H}_{f C_2{H}_2}^o=226.7 kJ/mol$?

b) What is the fuel value of acetylene, assuming the products are ${ CO}_2$ and $H_{2}O$ vapor?

Answer to Problem 9.111QA

Solution:

a) The standard heat of combustion of acetylene is $-1255.5\frac{kJ}{mol}$

b) The fuel value of acetylene is $48.22\frac{kJ}{g}$

Explanation of Solution

1) Concept:

Heat of combustion: The amount of energy released in the form of heat when the compound undergoes complete combustion in the presence of oxygen under standard condition.

For a balanced chemical equation, the standard heat of reaction is the sum of the ∆H⁰_f values of the products, each multiplied by its coefficient describing the reaction, and then subtracting the sum of the ∆H⁰_f values of the reactants, each multiplied by its coefficient.

Fuel value is the amount of energy generated by complete combustion of one gram of fuel

This can be calculated from the enthalpy change for combustion reaction.

2) Formula:

${∆H}_{rxn}^o=\sum n_{products }{∆H}_{f products}^o-\sum n_{reactants }{∆H}_{f reactants}^o$

where,

 $n_{products}$ is the number of moles of each product in the balanced chemical equation

$n_{reactants}$ is the number of moles of each reactant in the balanced chemical equation.

3) Given:

${∆H}_{f C_2{H}_2\left(g\right)}^o=226.7\frac{kJ}{mol}$

4) Calculation:

a) Heat of combustion of acetylene:

Equation for combustion of acetylene is

$C_2{H}_2\left(g\right)+O_2\left(g\right)\to {CO}_2\left(g\right)+H_{2}O\left(g\right)$

Taking an inventory of an atoms on both sides,

$2C+2H+2O\to 1C+2H+3O$

The H are already balanced.

To balance the C, we need to add the coefficient 2 in front of ${CO}_2.$

$C_2{H}_2\left(g\right)+O_2\left(g\right)\to 2{ CO}_2\left(g\right)+H_{2}O\left(g\right)$

Taking an inventory of atoms on both sides,

$2C+2H+2O\to 2C+2H+5O$

To balance the O, we need to add the coefficient $\frac{5}{2}$  in front of $O_2$.

$C_2{H}_2\left(g\right)+{\frac{5}{2}O}_2\left(g\right)\to 2{ CO}_2\left(g\right)+H_{2}O\left(g\right)$

To convert the fractional coefficient to whole numbers, we need to multiply each coefficient by 2.

$2C_2{H}_2\left(g\right)+{5O}_2\left(g\right)\to 4{ CO}_2\left(g\right)+{2 H}_{2}O\left(g\right)$

This is a balanced equation.

From appendix 4, the $∆H_f^0$ values are

Substances	$C_2{H}_2\left(g\right)$	$O_2\left(g\right)$	$C{O}_2\left(g\right)$	$H_{2}O \left(g\right)$
$∆H_f^0$  ($\frac{kJ}{mol}$)	+226.7	0.0	-393.5	-241.8

Inserting $∆H_f^0$ values for the products and reactants and their coefficients from the balanced

chemical equation into the formula,

${∆H}_{rxn}^0=Ʃn_{products}∆H_{f,products}^0-Ʃn_{reactants}∆H_{f,reactants}^0$

${∆H}_{rxn}^0=\left[\left(4 mol {CO}_2\right)\left(-393.5\frac{kJ}{mol}\right)+\left(2 mol{ H}_{2}O\right)\left(-241.8\frac{kJ}{mol}\right)\right] -\left[\left(2 mol {C_{2}H}_2\right)\left(+226.7\frac{kJ}{mol}\right)+\left(1 mol O_2\right)\left(0.0\frac{kJ}{mol}\right)\right]$

${∆H}_{rxn}^0 =\left[\left(-1574 kJ\right)+\left(-483.6 kJ\right)\right]-[\left(+453.4 kJ\right)+\left(0.0 kJ\right)]$

${∆H}_{rxn}^0 =\left[-1574 kJ-483.6 kJ\right]-[+453.4 kJ]$

${∆H}_{rxn}^0 = -2057.6 kJ-453.4 kJ$

${∆H}_{rxn}^0 =-2511 kJ$

This is the heat of reaction for 2 moles of acetylene.

Standard heat of combustion is expressed in units of $\frac{kJ}{mol}$.

So, standard heat of combustion of acetylene is

$\frac{-2511 kJ}{2 mol C_2{H}_2}= -1255.5\frac{kJ}{mol}$

Thus, the standard heat of combustion of acetylene is $-1255.5\frac{kJ}{mol}$

b) Fuel value of acetylene:

Heat of combustion of $C_2{H}_2$ is $-1255.5 kJ/mol$.

So, the mass of 1 mol of $C_2{H}_2=26.038 g$

Therefore, the fuel value of $C_2{H}_2$ per gram is

$Fuel value = \frac{1255.5 kJ}{mol}\times \frac{1 mol C_6{H}_{14} }{26.038 g}=48.22\frac{kJ}{g}$

Conclusion:

Heat of combustion is calculated using the standard heat of formation, and then fuel value is calculated.