Question

Consider the following cases of standard enthalpy of reaction (\( \Delta H_f^\circ \) in kJ mol\(^{-1}\)): \[ \text{C}_2\text{H}_6(g) + 7 \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + 3 \text{H}_2\text{O}(l) \quad \Delta H_1^\circ = -1550 \] \[ \text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2^\circ = -393.5 \] \[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2\text{O}(g) \quad \Delta H_3^\circ = -286 \] The magnitude of \( \Delta H_f^\circ \) of \( \text{C}_2\text{H}_6(g) \) is \(\_\_\_\_\_\) kJ mol\(^{-1}\) (Nearest integer). 

Hint:

Hess's law allows us to calculate the enthalpy change for a reaction by adding the enthalpy changes of individual steps, provided the reactions are appropriately manipulated.

Answer 1

Correct Answer: 84

Hess's law enables the calculation of \( \Delta H_f^\circ \) for \( \text{C}_2\text{H}_6(g) \).

The provided reaction is: \[ \text{C}_2\text{H}_6(g) + 7 \text{O}_2(g) \rightarrow 2 \text{CO}_2(g) + 3 \text{H}_2\text{O}(l) \quad \Delta H_1^\circ = -1550 \]

Summing the reactions and incorporating the enthalpy values for \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) yields: \[ \Delta H_f^\circ = -1550 + 2 \times 393.5 + 3 \times 286 = -84 \]

Consequently, the \( \Delta H_f^\circ \) for \( \text{C}_2\text{H}_6(g) \) is \( -84 \) kJ/mol.