We can calculate \( \Delta H_f^\circ \) for \( \text{C}_2\text{H}_6(g) \) using Hess's law.
Using the given reactions: \[ \text{C}_2\text{H}_6(g) + 7 \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + 3 \text{H}_2\text{O}(l) \quad \Delta H_1^\circ = -1550 \]
Adding the reactions and using the enthalpy values for \( \text{CO}_2 \) and \( \text{H}_2\text{O} \): \[ \Delta H_f^\circ = -1550 + 2 \times 393.5 + 3 \times 286 = -84 \]
Thus, the magnitude of \( \Delta H_f^\circ \) for \( \text{C}_2\text{H}_6(g) \) is \( -84 \) kJ/mol.

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: