Question

Isobutane (\(C_4H_{10}\)) is burnt completely in pure oxygen as per the reaction given below. Given that the standard heats of formation (in kcal/mole) of isobutane, carbon dioxide, and water vapour are -31.489, -94.052, and -60.150, respectively, the heat of reaction is ________ kcal (rounded off to 2 decimal places).
\[ C_4H_{10} + 6.5\,O_2 \to 4\,CO_2 + 5\,H_2O \]

Hint:

Use Hess's law: heat of reaction = sum of heats of formation of products minus reactants, weighted by stoichiometric coefficients. O2 has zero heat of formation.

Answer 1

Correct Answer: -645.47

Step 1: Build a two-step enthalpy cycle.
Instead of applying the formula directly, picture the reaction happening in two imaginary steps: first, isobutane is broken down completely into its elements (carbon and hydrogen); second, those same elements (together with the oxygen) are combined to form the products, $CO_2$ and $H_2O$. The total enthalpy change is the same by either route, since enthalpy is a state function.

Step 2: Enthalpy to break isobutane into its elements.
Forming 1 mole of isobutane from its elements releases $31.489$ kcal (since $\Delta H_f = -31.489$ kcal/mole), so breaking it back down into elements absorbs the same amount:
$\Delta H_1 = +31.489$ kcal.

Step 3: Enthalpy to form the products from the elements.
Forming 4 moles of $CO_2$ releases $4 \times 94.052 = 376.208$ kcal, and forming 5 moles of $H_2O$ vapour releases $5 \times 60.150 = 300.750$ kcal. Oxygen needs no formation step since it starts as an element.
$\Delta H_2 = -(376.208 + 300.750) = -676.958$ kcal.

Step 4: Add the two steps of the cycle.
$\Delta H_{rxn} = \Delta H_1 + \Delta H_2 = 31.489 - 676.958 = -645.469$ kcal.

Final Answer:
Rounded to 2 decimal places, $\Delta H_{rxn} \approx -645.47$ kcal, the same result as the direct Hess's law formula.
\[ \boxed{\Delta H_{rxn} \approx -645.47\ \text{kcal}} \]