Question

When $2\,$moles of ${C_2}{H_6}$ (g) are completely burnt $3120kJ$ of heat is liberated. The enthalpy of formation of ${C_2}{H_6}$ in $kJ/mol$is $X$. Find $X$ .
Given ; $\Delta {{\rm H}_f}$ for $C{O_2}$ (g) and ${H_2}O$ (l) are $- 395kJ$ and $- 285kJ$respectively.

Answer 1

The Heat of Reaction (also known and Enthalpy of Reaction) is that the change within the enthalpy of a chemical action that happens at a relentless pressure. It's a thermodynamic unit of measurement useful for calculating the quantity of energy per mole either released or produced in a very reaction. Since enthalpy comes from pressure, volume, and internal energy, all of which are state functions, enthalpy is additionally a state function.

Complete step by step answer:
Given:

Heat liberated $\,\left( {\Delta H^\circ } \right){{ }} = {{ }}3120{{ }}KJ\,$

$\,\Delta H{{ }}of\;C{O_2}\; = {{ }} - 395{{ }}KJ\,$

$\,\Delta H{{ }}\;of \;{H_2}O\; = {{ }} - 285{{ }}KJ\,$

To find:

$\,\Delta H{{ }}of\;{C_2}{H_6} = X\;\,$

The Formula to be used is:

$\Delta {{\rm H}^ \circ } = \Delta {H_{f(products)}}\; - \Delta {H_{f(reactant)}}$

$\Delta {{\rm H}^ \circ } =$ represents a change in the enthalpy of the reaction $\,(\Delta {H_{products}}\; - \Delta {H_{reactants}})\,$

a positive value indicates that the products have greater enthalpy, or that it is an endothermic reaction (heat is required).a negative value indicates that the reactants have greater enthalpy, or that it is an exothermic reaction (heat is produced) equal to signifies that the reaction is a standard enthalpy change, and occurs at a preset pressure/temperature.

Chemical equation:

The balanced chemical reaction involved in the combustion of two moles of ethane will be as follows:

$2{C_2}{H_6} + 7{O_2} \to 4C{O_2} + 6{H_2}O$

From the above-given chemical equation it can be interpreted that for combustion of $2$moles of ethane, $7$ moles of oxygen is required.The combustion yields $4$ moles of carbon dioxide and $6$ moles of water.

Calculation:

$\Delta {{\rm H}^ \circ } = \Delta {H_{f(products)}}\; - \Delta {H_{f(reactant)}}$

$3120 = \left[ {4 \times \Delta {H_{(C{O_2})}} + 6 \times \Delta {{\rm H}_{({H_2}O)}}} \right] - \left[ {2 \times \Delta {{\rm H}_{({C_2}{H_6})}} + 7 \times \Delta {{\rm H}_{({O_2})}}} \right]$

Given, $\left[ {\Delta {{\rm H}_{({C_2}{H_6})}}} \right] = X$

$3120 = \left[ {4 \times ( - 395) + 6 \times ( - 285)} \right] - \left[ {2X + 7 \times 0} \right]$

On solving,

$\Rightarrow 2X = ( - 1580) + ( - 1710) - 3120$

$\Rightarrow 2X = - 170$

$\Rightarrow X = - 85$

Which means,

$\,\left[ {\Delta {{\rm H}_{({C_2}{H_6})}}} \right] = - 85kJ/mol\,$

Thus, the heat of formation of ${C_2}{H_6}$ is calculated as $- 85kJ/mol$.

Note: Enthalpy of formation ($\Delta {H_f}$ ) is that the enthalpy change for the formation of one mole of a compound from its component elements, like the formation of greenhouse gas from carbon and oxygen.
The magnitude of $\Delta {H_f}$ for a reaction depends on the physical states of the reactants and also the products (gas, liquid, solid, or solution), the pressure of any gases present, and therefore the temperature at which the reaction is dispensed.