Question

The heat of combustion of carbon, hydrogen and acetylene are -394kJ, -286kJ and -1301kJ respectively. Calculate heat of formation of ${C_2}{H_2}$.
A. 621 kJ
B. 454 kJ
C. -227 kJ
D. 227 kJ

Answer 1

Hint:Heat of formation is the heat change (either evolved or absorbed) accompanying the formation of 1 mole of a substance from its elements under a given condition of temperature and pressure. Also heat of combustion is the amount of heat liberated when one mole of substance is completely burnt or oxidized in air.

Complete step by step solution:
We are given the heat of combustion of carbon, hydrogen and acetylene and asked to find the heat of formation of acetylene (${C_2}{H_2}$). So since we are given the heat of formation, let’s find out what that really means.
The heat of combustion is heat liberated when it is completely burnt or oxidized in air. So let’s make carbon to oxidise in air.
$C + {O_2} \to C{O_2}$ --(1)
When carbon is oxidised in the presence of air we get carbon dioxide as the product of the heat of combustion, $\vartriangle H = - 394kJ$ which is given in the question.
Now, we can see what happens when hydrogen is oxidised in the presence of air.
${H_2} + \dfrac{1}{2}{O_2} \to {H_2}O$ --(2)
We get water when we oxidise hydrogen in the presence of air and the heat of combustion is $\vartriangle H = - 286kJ$.
Now, let’s look at the combustion of acetylene,
${C_2}{H_2} + \dfrac{5}{2}{O_2} \to 2{\text{C}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}}$ --(3)
The equation is balanced for 1 mole of acetylene as the heat of combustion is given for one mole of substance. The heat of combustion of one mole of acetylene is $\vartriangle H = - 1301kJ$
We have to find the formation of acetylene, which is
$2C + {H_2} \to {C_2}{H_2}$
For solving these questions, first of all write the target equation which consists of the compounds for which heat of combustion has been taken. Then we can apply the formula
$\vartriangle {H_{reaction}} = \vartriangle {H_{C\left( {products} \right)}} - \vartriangle {H_{C\left( {reactants} \right)}}$
Now multiplying the reaction(1)
$2C + 2{O_2} \to 2C{O_2}$ $\vartriangle H = 2 \times - 394kJ = - 788kJ$ --(4)
Adding (4)+(2) and subtracting/reversing it with (3), we get our reaction to form acetylene.
$2C + 2{O_2} + {H_2} + \dfrac{1}{2}{O_2} + 2{\text{C}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}} \to C{O_2} + {H_2}O + {C_2}{H_2} + \dfrac{5}{2}{O_2}$
Now, after cancellation, we get
$2C + {H_2} \to {C_2}{H_2}$
Thus heat of reaction is
$\vartriangle {H_{reaction}} = - 788 + \left( { - 286} \right) + 1301$
$\vartriangle {H_{reaction}} = 227kJ$

Therefore, the heat of formation of acetylene is 227kJ.

Note:
If the heat of formation is greater than zero we call it an endothermic reaction and if it is less than zero we call it an exothermic reaction.
In other words,
If $\sum \vartriangle {H_P} > \sum \vartriangle {H_{R,\;}}\;\;\vartriangle H = + ve$ i.e., endothermic reaction
If $\sum \vartriangle {H_P} < \sum \vartriangle {H_{R,\;}}\;\;\vartriangle H = - ve$ i.e., exothermic reaction