The problem asks us to calculate the enthalpy of formation of $C_2H_6(g)$ given its enthalpy of combustion, the enthalpy of formation of $CO_2(g)$, and the enthalpy of formation of $H_2O(l)$. We are given: Enthalpy of combustion of $C_2H_6(g)$ = -1559.7 kJ/mol Enthalpy of formation of $CO_2(g)$ = -393.5 kJ/mol Enthalpy of formation of $H_2O(l)$ = -285.8 kJ/mol

1. Problem Description

The problem asks us to calculate the enthalpy of formation of $C_2H_6(g)$ given its enthalpy of combustion, the enthalpy of formation of $CO_2(g)$, and the enthalpy of formation of $H_2O(l)$. We are given:

Enthalpy of combustion of $C_2H_6(g)$ = -1559.7 kJ/mol

Enthalpy of formation of $CO_2(g)$ = -393.5 kJ/mol

Enthalpy of formation of $H_2O(l)$ = -285.8 kJ/mol

2. Solution Steps

First, write the balanced equation for the combustion of ethane ($C_2H_6$):

$C_2H_6(g) + \frac{7}{2}O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$

Next, write the equation for the standard enthalpy of combustion in terms of the enthalpies of formation:

$\Delta H_{combustion} = \sum n_p \Delta H_{f(products)} - \sum n_r \Delta H_{f(reactants)}$

Here,

$\Delta H_{combustion} = -1559.7$ kJ/mol

$\Delta H_{f(CO_2)} = -393.5$ kJ/mol

$\Delta H_{f(H_2O)} = -285.8$ kJ/mol

$\Delta H_{f(O_2)} = 0$ kJ/mol (since $O_2$ is in its standard state)

Substituting the known values into the equation:

$-1559.7 = [2 \times (-393.5) + 3 \times (-285.8)] - [1 \times \Delta H_{f(C_2H_6)} + \frac{7}{2} \times 0]$

$-1559.7 = [-787 - 857.4] - \Delta H_{f(C_2H_6)}$

$-1559.7 = -1644.4 - \Delta H_{f(C_2H_6)}$

$\Delta H_{f(C_2H_6)} = -1644.4 + 1559.7$

$\Delta H_{f(C_2H_6)} = -84.7$ kJ/mol

3. Final Answer

The enthalpy of formation of $C_2H_6(g)$ is -84.7 kJ/mol.

Therefore, the answer is (2).