Question

When 2 moles of ${C_2}{H_6}.$ . $\Delta {H_f}$ are completely burnt, $3129\,kJ$ of heat is liberated. Calculate the heat of formation of ${C_2}{H_6}.\Delta {H_f}$ for $C{O_2}$ and ${H_2}O$are $- 395\,kJ$and $- 286\,kJ$ respectively

Answer 1

Enthalpy is defined as the total heat content of the system at constant pressure. The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Complete step by step answer:
The equation for the combustion of ${C_2}{H_6}$ is:
$2{C_2}{H_5} + 7{O_2} \to 4C{O_2} + 6{H_2}O;\Delta {H^ \circ } = - 3129kJ$

We know,
$\Delta {H^ \circ } = \Delta H_{f(products)}^ \circ - \Delta H_{f(react.)}^ \circ$

Now we will put the given values in the equation
$\Rightarrow - 3129 = \left[ {4 \times \Delta H_{(C{O_2})}^ \circ + 6 \times \Delta H_{({H_2}O)}^ \circ } \right] - \left[ {2 \times \Delta H_{({C_2}{H_6})}^ \circ + 7 \times 0} \right]$

(We know the standard heat values of $C{O_2}$ and ${H_2}O$ so we will put the values in the equation. Since oxygen is in elemental state its heat of formation will be 0)
$\Rightarrow - 3129 = \left[ {4 \times ( - 395) + 6 \times ( - 286)} \right] - \left[ {2 \times \Delta H_{({C_2}{H_6})}^ \circ + 7 \times 0} \right]$
$\Rightarrow - 3129 = \left[ { - 1580 - 1716} \right] - 2 \times \Delta H_{({C_2}{H_6})}^ \circ$
$\Rightarrow 2 \times \Delta H_{({C_2}{H_6})}^ \circ = - 3296 + 3129$
$\Rightarrow 2 \times \Delta H_{({C_2}{H_6})}^ \circ = - 167$
$\Rightarrow \Delta H_{({C_2}{H_6})}^ \circ = \dfrac{{ - 167}}{2}$
$\Rightarrow \Delta H_{({C_2}{H_6})}^ \circ = - 83.5\,kJ$

After calculation we will get value of heat of formation of ${C_2}{H_6}.\Delta {H_f}$ is $- 83.5KJ$

Additional Information:
Enthalpy may also be defined as the sum of the internal energy (U) and pressure-volume energy (PV) of a system.
Mathematically,

$H = U + PV$

Note: Key points for doing enthalpy calculations.

1.When a reaction is reversed, the value of $\Delta H$ stays the same, but the sign changes.
2.When the balanced chemical equation for the reaction is multiplied by an integer, the corresponding value of $\Delta H$ must be multiplied by the integer as well.
3.The change in enthalpy for a reaction can be calculated from the enthalpies of formation of the reactants and the products.
4.Elements in their standard states do not contribute to the enthalpy calculations for the reactions, since the enthalpy of an element in its standard state is zero.
5.Allotropes of an element other than the standard state generally have non-zero standard enthalpies of formation.