At 25°C and 1 atm pressure,the enthalpies of combustion are as given below:
$Substance$ $H_2$ $C(graphite)$ $C_2H_6(g)$
$\frac{D_cH^{\theta}}{kJmol^{-1}}$ $-286.0$ $-394.0$ -1560.0
The enthalpy of formation of ethane is

Question

The group 14 elements A and B have the first ionisation enthalpy values of 708 and 715 kJ mol$^{-1}$ respectively. The above values are lowest among their group members. The nature of their ions A$^{2+}$ and B$^{4+}$ respectively is:

Answer 1

To find the enthalpy of formation of ethane $(C_2H_6)$, we will first write the combustion and formation reactions and then use the given enthalpy changes for combustion to determine the desired formation enthalpy.

The formation reaction for ethane $(C_2H_6(g))$ from its elements in their standard states is:

2C(graphite) + 3H_2(g) \rightarrow C_2H_6(g)

The combustion reactions and their respective enthalpies provided in the question are:

H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)\quad \Delta_cH^\theta = -286.0\ kJ/mol
C(graphite) + O_2(g) \rightarrow CO_2(g)\quad \Delta_cH^\theta = -394.0\ kJ/mol
C_2H_6(g) + \frac{7}{2}O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)\quad \Delta_cH^\theta = -1560.0\ kJ/mol

Now, the enthalpy change of formation of ethane can be calculated using Hess's Law which states that enthalpy change for a reaction is the same whether it occurs in one step or several steps. We can break down the combustion of ethane into formation parts and then calculate:

The enthalpy change for formation can be expressed as:

\Delta_fH^\theta(C_2H_6) = \text{[enthalpy of combustion of elements]} - \text{[enthalpy of combustion of compound]}

Substituting the given values:

\Delta_fH^\theta(C_2H_6) = \left[2 \cdot (-394.0)\ + 3 \cdot \left(-286.0\right)\right] - (-1560.0)

\Delta_fH^\theta(C_2H_6) = \left(-788.0 - 858.0\right) + 1560.0

\Delta_fH^\theta(C_2H_6) = -1646.0 + 1560.0 = -86.0\ kJ/mol

Thus, the enthalpy of formation of ethane is -86.0 kJ mol^-1.

Among the given options, the correct answer is: -86.0 kJ mol^-1

Answer 2