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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...

The enthalpies of the formation for C2H4(g),CO2(g){C_2}{H_4}(g),C{O_2}(g) and H2O(l){H_2}O(l) at 25∘C{25^ \circ }C and 1atm1 atm pressure be 52,−39452, - 394 and −286kJ/mol - 286kJ/mol respectively. Find the enthalpy of combustion of C2H4(g){C_2}{H_4}(g).
A.−141.2kJ/mol - 141.2kJ/mol
B.−1412kJ/mol - 1412kJ/mol
C.+141.2kJ/mol + 141.2kJ/mol
D.+1412kJ/mol + 1412kJ/mol

Explanation

Solution

Write the equations of the formation of C2H4(g),CO2(g){C_2}{H_4}(g),C{O_2}(g) and H2O(l){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 C2H4(g){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, C2H4(g),CO2(g){C_2}{H_4}(g),C{O_2}(g) and H2O(l){H_2}O(l) are formed at 25∘C{25^ \circ }C and 1atm1atm pressure and their enthalpies of formation are 52,−39452, - 394 and −286kJ/mol - 286kJ/mol respectively. So, the equations of formation of C2H4(g),CO2(g){C_2}{H_4}(g),C{O_2}(g) and H2O(l){H_2}O(l) are –
2C(g)+2H2(g)→C2H4(g);ΔH1=52kJ/mol⋯(1) C(g)+O2(g)→CO2(g);ΔH2=−394kJ/mol⋯(2) H2(g)+12O2(g)→H2O(g);ΔH3=−286kJ/mol⋯(3)  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)+32O2(g)+H2(g)→CO2+H2O;ΔH4=−680kJ/mol⋯(4)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 22, we get –
2C(g)+3O2(g)+2H2(g)→2CO2+2H2O;ΔH5=−1360kJ/mol⋯(5)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 –
C2H4(g)+3O2→2CO2+2H2O;ΔH6=−1412kJ/mol⋯(6){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 C2H4{C_2}{H_4}.
Therefore, the enthalpy of combustion we got in the equation (6) is −1412kJ/mol - 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.