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

Step 1: Know what enthalpy of combustion means.
The enthalpy of combustion is the heat change when one mole of a fuel burns fully in oxygen. We find it with Hess's law: \[ \Delta H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants}). \]
Step 2: Write the balanced burning equation.
Methane burns to give carbon dioxide and liquid water: \[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l). \] Note that two moles of water form.
Step 3: List the formation enthalpies.
Given $\Delta_f H(CH_4) = -x$, $\Delta_f H(CO_2) = -y$, $\Delta_f H(H_2O) = -z$. An element in its standard form, $O_2$, has $\Delta_f H = 0$.
Step 4: Build the products minus reactants sum.
\[ \Delta_c H = \big[\Delta_f H(CO_2) + 2\Delta_f H(H_2O)\big] - \big[\Delta_f H(CH_4) + 2\Delta_f H(O_2)\big]. \]
Step 5: Substitute the symbols.
\[ \Delta_c H = \big[(-y) + 2(-z)\big] - \big[(-x) + 0\big] = -y - 2z + x. \]
Step 6: Tidy the final form.
Taking a negative sign common from the first part gives \[ \boxed{\Delta_c H = -(y + 2z - x)} \]