Question

The enthalpies of the formation for ${C_2}{H_4}(g),C{O_2}(g)$ and ${H_2}O(l)$ at ${25^ \circ }C$ and $1 atm$ pressure be $52, - 394$ and $ - 286kJ/mol$ respectively. Find the enthalpy of combustion of ${C_2}{H_4}(g)$.
A.$ - 141.2kJ/mol$
B.$ - 1412kJ/mol$
C.$ + 141.2kJ/mol$
D.$ + 1412kJ/mol$

Answer 1

Hint: Write the equations of the formation of ${C_2}{H_4}(g),C{O_2}(g)$ and ${H_2}O(l)$ with their enthalpies involved during their formation. Then, by solving all those three equations, find an equation for the combustion of ${C_2}{H_4}(g)$ and then the enthalpy.

Complete Step by Step Solution:
We know that
Enthalpy of combustion refers to the reaction where one mole of the compound in a reaction completely combines with oxygen gas. Unlike the formation reactions, combustion is referring to one of the reactants not the products.
It can be also defined as the energy released when 1 mole of a compound is burned in excess of oxygen to form products at 298k and 1 atm pressure or under standard conditions.
In the question, it is given that, ${C_2}{H_4}(g),C{O_2}(g)$ and ${H_2}O(l)$ are formed at ${25^ \circ }C$ and $1atm$ pressure and their enthalpies of formation are $52, - 394$ and $ - 286kJ/mol$ respectively. So, the equations of formation of ${C_2}{H_4}(g),C{O_2}(g)$ and ${H_2}O(l)$ are –
$
  2C(g) + 2{H_2}(g)\xrightarrow{{}}{C_2}{H_4}(g);\Delta {H_1} = 52kJ/mol \cdots \left( 1 \right) \\
  C(g) + {O_2}(g)\xrightarrow{{}}C{O_2}(g);\Delta {H_2} = - 394kJ/mol \cdots \left( 2 \right) \\
  {H_2}(g) + \dfrac{1}{2}{O_2}(g)\xrightarrow{{}}{H_2}O(g);\Delta {H_3} = - 286kJ/mol \cdots \left( 3 \right) \\
 $
The chemical equations written above are all balanced chemical equations.
Now, adding equation (2) and equation (3), we get –
$C(g) + \dfrac{3}{2}{O_2}(g) + {H_2}(g)\xrightarrow{{}}C{O_2} + {H_2}O;\Delta {H_4} = - 680kJ/mol \cdots \left( 4 \right)$
Multiplying equation (4), by $2$, we get –
$2C(g) + 3{O_2}(g) + 2{H_2}(g)\xrightarrow{{}}2C{O_2} + 2{H_2}O;\Delta {H_5} = - 1360kJ/mol \cdots \left( 5 \right)$
Now, subtracting equation (5) from equation (1), we get –
${C_2}{H_4}(g) + 3{O_2}\xrightarrow{{}}2C{O_2} + 2{H_2}O;\Delta {H_6} = - 1412kJ/mol \cdots \left( 6 \right)$
Equation (6) is the required equation. It is the combustion reaction of ${C_2}{H_4}$.
Therefore, the enthalpy of combustion we got in the equation (6) is $ - 1412kJ/mol$.

Hence, the correct option is (B).

Note: Enthalpy change of atomization is defined as the enthalpy change which accompanies the formation of one mole of gaseous atoms from its element in its standard state (either solid/liq/gas). It is an endothermic reaction.
Lattice enthalpy is the energy required to break a solid ionic compound to its constituent gaseous positive and negative ions.