Questions: Using the bond energies found in the table below, show the calculation of the approximate enthalpy change, ΔH, for the following reaction: C3H8(g)+5O2(g) → 3CO2(g)+4H2O(l) Bond Bond Energy (kJ/mol) C-H 415 C-C 345 O=O 498 C=O 741 H-O 464

Using the bond energies found in the table below, show the calculation of the approximate enthalpy change, ΔH, for the following reaction:
C3H8(g)+5O2(g) → 3CO2(g)+4H2O(l)
Bond  Bond Energy (kJ/mol)
C-H  415
C-C  345
O=O  498
C=O  741
H-O  464
Transcript text: Using the bond energies found in the table below, show the calculation of the approximate enthalpy change, $\Delta H$, for the following reaction: \[ \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \] \begin{tabular}{|l|l|} \hline Bond & Bond Energy $\left(\frac{\mathrm{kJ}}{\mathrm{mol}}\right)$ \\ \hline C-H & 415 \\ \hline C-C & 345 \\ \hline O=O & 498 \\ \hline C=O & 741 \\ \hline H-O & 464 \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: Calculate the total bond energy of the reactants.

C3H8 has two C-C bonds and eight C-H bonds. O2 has five O=O bonds. Total bond energy of reactants = 2(345) + 8(415) + 5(498) = 690 + 3320 + 2490 = 6500 kJ/mol

Step 2: Calculate the total bond energy of the products.

CO2 has three molecules, each with two C=O bonds. H2O has four molecules, each with two H-O bonds. Total bond energy of products = 3 * 2 * (741) + 4 * 2 * (464) = 4446 + 3712 = 8158 kJ/mol

Step 3: Calculate the enthalpy change.

ΔH = (Total bond energy of reactants) - (Total bond energy of products) ΔH = 6500 kJ/mol - 8158 kJ/mol = -1658 kJ/mol

Final Answer

-1658 kJ/mol

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