Question

The standard enthalpies of formation of $\mathrm{CH_4(g)}$, $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$ are $-78$, $-395$ and $-288\ \mathrm{kJ\ mol^{-1}}$ respectively. Calculate $\Delta H$ for: \[ \mathrm{2CH_4(g) + 4O_2(g) \rightarrow 2CO_2(g) + 4H_2O(l)} \] }

• A. $+890.3$
• B. $-890.3$
• C. $-1780.6$
• D. $+1780.6$

Hint:

Always remember: \[ \Delta H = \text{Products} - \text{Reactants} \] A negative value of $\Delta H$ indicates that the reaction is exothermic.

Answer 1

The Correct Option is C

Understanding the Concept: The enthalpy change of reaction is calculated using: \[ \Delta H_{rxn} = \sum \Delta H_f^\circ(\text{Products}) - \sum \Delta H_f^\circ(\text{Reactants}) \] Standard enthalpy of formation of elemental oxygen: \[ \Delta H_f^\circ(O_2)=0 \] because oxygen exists in standard state.
Step 1: Write Formation Enthalpies} Products: \[ \mathrm{CO_2(g)} = -393.5\ \mathrm{kJ\ mol^{-1}} \] \[ \mathrm{H_2O(l)} = -285.8\ \mathrm{kJ\ mol^{-1}} \] Reactant: \[ \mathrm{CH_4(g)} = -74.8\ \mathrm{kJ\ mol^{-1}} \]
Step 2: Calculate Total Enthalpy of Products} There are: \[ 2\ mol\ of\ CO_2 \] and \[ 4\ mol\ of\ H_2O \] Thus: \[ 2(-393.5) + 4(-285.8) \] \[ = -787 -1143.2 \] \[ = -1930.2\ \mathrm{kJ} \]
Step 3: Calculate Total Enthalpy of Reactants} There are: \[ 2\ mol\ of\ CH_4 \] Thus: \[ 2(-74.8) + 4(0) \] \[ = -149.6\ \mathrm{kJ} \]
Step 4: Find Enthalpy Change} \[ \Delta H = -1930.2 -(-149.6) \] \[ \Delta H = -1930.2 +149.6 \] \[ \Delta H = -1780.6\ \mathrm{kJ} \] Hence, the correct answer is: \[ \boxed{(C)\ -1780.6} \]