Question

Given that the standard enthalpy of combustion of \(C(s)\) and \(CS_2(l)\) are \(-393.3\) and \(-1108.76 \, \text{kJ/mol}\) respectively and the standard enthalpy of formation of \(CS_2\) is \(128.02 \, \text{kJ/mol}\). What is \(\Delta H_f^\circ\) of \(SO_2\)?

• A. \(-510.6 \, \text{kJ/mol}\)
• B. \(-293.72 \, \text{kJ/mol}\)
• C. \(-321.2 \, \text{kJ/mol}\)
• D. \(-587 \, \text{kJ/mol}\)

Hint:

Use Hess's law carefully: write combustion reactions and apply \(\Delta H = \text{products} - \text{reactants}\). Always substitute standard enthalpy of formation values correctly.

Answer 1

The Correct Option is B

Step 1: Write combustion reactions.
For carbon:
\[ C(s) + O_2(g) \rightarrow CO_2(g), \quad \Delta H = -393.3 \, \text{kJ/mol} \] For \(CS_2\):
\[ CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g), \quad \Delta H = -1108.76 \, \text{kJ/mol} \]

Step 2: Write formation reaction of \(CS_2\).
\[ C(s) + 2S(s) \rightarrow CS_2(l), \quad \Delta H_f^\circ = +128.02 \, \text{kJ/mol} \]

Step 3: Apply Hess's Law.
We express combustion of \(CS_2\) using enthalpies of formation:
\[ \Delta H = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \] \[ -1108.76 = [\Delta H_f^\circ (CO_2) + 2\Delta H_f^\circ (SO_2)] - [\Delta H_f^\circ (CS_2)] \]

Step 4: Substitute known values.
\[ \Delta H_f^\circ (CO_2) = -393.3 \, \text{kJ/mol} \] \[ \Delta H_f^\circ (CS_2) = +128.02 \, \text{kJ/mol} \] Substitute:
\[ -1108.76 = [-393.3 + 2\Delta H_f^\circ (SO_2)] - 128.02 \]

Step 5: Simplify the equation.
\[ -1108.76 = -393.3 - 128.02 + 2\Delta H_f^\circ (SO_2) \] \[ -1108.76 = -521.32 + 2\Delta H_f^\circ (SO_2) \]

Step 6: Solve for \(\Delta H_f^\circ (SO_2)\).
\[ -1108.76 + 521.32 = 2\Delta H_f^\circ (SO_2) \] \[ -587.44 = 2\Delta H_f^\circ (SO_2) \] \[ \Delta H_f^\circ (SO_2) = \frac{-587.44}{2} = -293.72 \, \text{kJ/mol} \]

Step 7: Final Answer.
\[ \boxed{-293.72 \, \text{kJ/mol}} \] Hence, the correct answer is option (B).