Question

Combustion of methane gives \(CO_2(g)\) and \(H_2O(l)\). What is enthalpy of combustion \((\Delta_c H^\circ \text{ in kJ mol}^{-1})\) of \(CH_4(g)\) at \(298\,K\)?
\[ (\Delta_f H^\circ(CH_4(g)) = -x\,\text{kJ mol}^{-1},\; \Delta_f H^\circ(CO_2(g)) = -y\,\text{kJ mol}^{-1},\; \Delta_f H^\circ(H_2O(l)) = -z\,\text{kJ mol}^{-1}) \]

• A. \(-(y + 2z - x)\)
• B. \((2z - y + x)\)
• C. \((2z + x - y)\)
• D. \(-(2y + z + x)\)

Hint:

Always remember: \[ \Delta H^\circ = \text{Products} - \text{Reactants} \] using standard enthalpies of formation. Also: \[ \Delta_f H^\circ(O_2)=0 \] because oxygen is in its standard elemental state.

Answer 1

The Correct Option is A

Concept: The standard enthalpy of reaction is calculated using: \[ \Delta H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}) \] This relation is derived from Hess's law. For combustion reactions:

• Products are usually \(CO_2\) and \(H_2O\)

• Oxygen gas has zero standard enthalpy of formation

Step 1: Write the balanced combustion equation. Combustion of methane: \[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \]

Step 2: Apply enthalpy of formation formula. Using: \[ \Delta_c H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}) \] Thus, \[ \Delta_c H^\circ = \left[ \Delta_f H^\circ(CO_2) + 2\Delta_f H^\circ(H_2O) \right] - \left[ \Delta_f H^\circ(CH_4) + 2\Delta_f H^\circ(O_2) \right] \]

Step 3: Substitute the given values. We know: \[ \Delta_f H^\circ(CO_2) = -y \] \[ \Delta_f H^\circ(H_2O) = -z \] \[ \Delta_f H^\circ(CH_4) = -x \] and \[ \Delta_f H^\circ(O_2)=0 \] Substituting: \[ \Delta_c H^\circ = [(-y)+2(-z)]-[(-x)+0] \] \[ = -y-2z+x \] Taking negative common: \[ = -(y+2z-x) \]

Step 4: Write the final answer. Hence, \[ \boxed{\Delta_c H^\circ = -(y+2z-x)} \] Therefore, the correct option is: \[ \boxed{(A)} \]