Question

Calculate the standard heat of formation of $C_{10}{H}_8$ (naphthalene) if standard heat of combustion of naphthalene is -1231.0 Kcal at 298 K and standard heats of formation of ${CO}_2$ (g) and $H_{2}O$ (l) are -94.0 Kcal and -68.4 Kcal respectively.
(a) 14.4 Kcal
(b) 17.4 Kcal
(c) 16.4 Kcal
(d) 15.4 Kcal

Answer 1

Hint: To solve this question, write down the chemical reaction for the formation of water, formation of carbon dioxide and combustion of naphthalene. Then use the Hess’ law in order to calculate the enthalpy of formation of naphthalene.

Complete step by step solution:
Let us first understand the meaning of standard heat of combustion and standard heat of formation. The standard heat of combustion or standard enthalpy of combustion for a substance is the heat change when one mole of the substance is burnt completely and forms products at standard conditions i.e. a temperature of 298 K and a pressure of 1 bar. It is denoted by the symbol ${\mathrm{\Delta }}_c{H}^o$.
The standard heat of formation or standard enthalpy of formation for a substance is the heat change accompanied when one mole of a substance in its standard state is formed from its elements that are also in their standard state i.e. at a temperature of 298 K and 1 bar pressure. It is denoted by the symbol ${\mathrm{\Delta }}_f{H}^o$.
Now let us solve the question.
Naphthalene ($C_{10}{H}_8$) on complete combustion gives carbon dioxide and water. The reaction is given below:
$C_{10}{H}_8(s)+{12O}_2(g)\stackrel{combustion}{\to }{10CO}_2(g)+{4H}_{2}O(l)$….(1)
The standard enthalpy of combustion for naphthalene is ${\mathrm{\Delta }}_c{H}^o$ =-1231.0 Kcal.
Now it is given that the standard heat of formation of carbon dioxide is -94.0 Kcal. The reaction is given below:
$C(s)+O_2(g)\to {CO}_2(g)$….(2)
The standard enthalpy of formation of water is -68.4 Kcal. The reaction is given below:
$H_2(g)+{\frac{1}{2}O}_2(g)\to H_{2}O(l)$….(3)
Multiplying equation (2) by 10 and equation (3) by 4 and adding the two we will get:
$\begin{array}{c}Equation(2):C(s)+O_2(g)\to {CO}_2(g)]\times 10\\ {Equation(3):H}_2(g)+{\frac{1}{2}O}_2(g)\to H_{2}O(l)]\times 4\\ Overallequation(4):10C(s)+12O_2(g)+{4H}_2(g)\to 10{CO}_2(g)+4H_{2}O(g)\end{array}$
Therefore the heat of reaction for equation (4) is: $10\times (-94.0Kcal)+[4\times (-68.4Kcal)]$= -1213.6 Kcal.
Now, subtracting equation (1) from equation (4) we will get the equation (5) for the formation of naphthalene.
$\begin{array}{c}Equation(4):10C(s)+12O_2(g)+4H_2(g)\to 10{CO}_2(g)+4H_{2}O(l)\\ Equation(1):C_{10}{H}_8(s)+12O_2(g)\to 10{CO}_2(g)+4H_{2}O(l)]\times (-1)\\ Overallequation(5):10C(s)+{4H}_2(g)\to C_{10}{H}_8(s)\end{array}$
The enthalpy of formation for reaction (5) will be: -1213.6 Kcal – (-1231.0 Kcal) = +17.4 Kcal.

Hence the standard heat of formation of naphthalene is (b) 17.4 Kcal.

Note: The standard conditions of 298 K temperature and 1 bar pressure are applicable for any enthalpy change that is taking place at standard conditions. Also when we are considering the standard enthalpy of combustion, it is not the combustion that is taking place at 298 K since when a substance burns, its temperature will be higher than 298 K; rather, it is the heat change that is taking place such that the initial reactants and the final products are at 298 K.