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).
\[ C_4H_{10} + 6.5\,O_2 \to 4\,CO_2 + 5\,H_2O \]
\[ C_4H_{10} + 6.5\,O_2 \to 4\,CO_2 + 5\,H_2O \]
Show Hint
Correct Answer: -645.47
Solution and Explanation
For any reaction, the heat of reaction is the sum of the standard heats of formation of the products minus the sum of the standard heats of formation of the reactants, each multiplied by its stoichiometric coefficient.
\[ \Delta H_{rxn} = \sum \Delta H_{f,products} - \sum \Delta H_{f,reactants} \]
Oxygen (\(O_2\)) is an element in its standard state, so its heat of formation is zero.
Step 2: List the given values and coefficients.
Reactant: 1 mole isobutane, \(\Delta H_f = -31.489\) kcal/mole.
Products: 4 moles \(CO_2\), \(\Delta H_f = -94.052\) kcal/mole each; 5 moles \(H_2O\) (vapour), \(\Delta H_f = -60.150\) kcal/mole each.
Step 3: Sum the heats of formation of the products.
\[ \Delta H_{f,products} = 4(-94.052) + 5(-60.150) = -376.208 - 300.750 = -676.958\ \text{kcal} \]
Step 4: Sum the heats of formation of the reactants.
\[ \Delta H_{f,reactants} = 1(-31.489) + 6.5(0) = -31.489\ \text{kcal} \]
Step 5: Apply Hess's law.
\[ \Delta H_{rxn} = -676.958 - (-31.489) = -645.469\ \text{kcal} \]
Final Answer:
Rounded to 2 decimal places, the heat of reaction is about -645.47 kcal (the negative sign shows the reaction is exothermic, it releases heat).
\[ \boxed{\Delta H_{rxn} \approx -645.47\ \text{kcal}} \]
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