Question

Calculate \(\Delta _{f}H\) for \(\text{C}(s)+2\text{H}_{2}(g)\rightarrow \text{CH}_{4}(g)\)
i. \(\text{C}(s)+\text{O}_{2}(g)\rightarrow \text{CO}_{2}(g);\Delta H_{1}=-393.5\) kJ/mol
ii. \(\text{H}_{2}(g)+1/2\text{O}_{2}(g)\rightarrow \text{H}_{2}\text{O}(l);\Delta H_{2}=-285.8\) kJ/mol
iii. \(\text{CH}_{4}(g)+2\text{O}_{2}(g)\rightarrow \text{CO}_{2}(g)+2\text{H}_{2}\text{O}(l);\Delta H_{3}=-890.3\) kJ/mol

• A. \(-1855.4\) kJ/mol
• B. \(-211\) kJ/mol
• C. \(-74.8\) kJ/mol
• D. \(1068.4\) kJ/mol

Hint:

Combine the three equations so that they add up to the formation of methane, then combine the enthalpies the same way.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
By Hess's law the total enthalpy change depends only on the start and end states, so we can add the enthalpies of equations that add up to the target.

Step 2: Combine the Equations
Target: \(\text{C}(s)+2\text{H}_2(g)\rightarrow\text{CH}_4(g)\).
Take (i) as it is, take (ii) twice, and reverse (iii).
\(\Delta_fH=\Delta H_1+2\Delta H_2-\Delta H_3\)

Step 3: Calculation
\(\Delta_fH=-393.5+2(-285.8)-(-890.3)\)
\(=-393.5-571.6+890.3=-74.8\) kJ/mol

Step 4: Compare the options
The result matches option (C).

Final Answer:
The enthalpy of formation of methane is minus 74.8 kJ per mole. \[ \boxed{\text{(C) }\text{-74.8 kJ/mol}} \]