Question

What is $\Delta_r H^\ominus$ (in kJ mol$^{-1}$) for the following reaction at 298 K? $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$ (Given: $\Delta_f H^\ominus$ of $C_3H_8(g)$, $CO_2(g)$ and $H_2O(l)$ is -104, -393 and -285 kJ mol$^{-1}$ respectively)

• A. +2215
• B. -2215
• C. -2427
• D. -2323

Hint:

Remember to multiply by stoichiometric coefficients and define $\Delta_f H^\ominus$ of $O_2$ as 0.

Answer 1

The Correct Option is B

Step 1: Recall the Hess-type formula.
The reaction enthalpy equals the total formation enthalpy of products minus that of reactants: \[ \Delta_r H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants}). \]

Step 2: List the given data.
The values are $\Delta_f H^\ominus$: $C_3H_8 = -104$, $CO_2 = -393$, $H_2O(l) = -285$ kJ/mol. Oxygen as an element has $\Delta_f H^\ominus = 0$.

Step 3: Add up the products.
There are 3 $CO_2$ and 4 $H_2O$: \[ 3(-393) + 4(-285) = -1179 - 1140 = -2319 \text{ kJ}. \]

Step 4: Add up the reactants.
One $C_3H_8$ plus 5 $O_2$: \[ -104 + 5(0) = -104 \text{ kJ}. \]

Step 5: Subtract.
\[ \Delta_r H^\ominus = -2319 - (-104) = -2319 + 104 = -2215 \text{ kJ/mol}. \]

Step 6: State the answer.
The negative sign shows heat is released, as expected for burning a fuel.
\[ \boxed{-2215} \]