Question

pq-13. the lewis dot structure of ethanol is given below. estimate the enthalpy of combustion of ethanol (c₂h₅oh) using the bond enthalpies in the table. \\(\ce{c2h5oh + 3o2 -> 2co2 + 3h2o}\\) \\(\

$$\begin{array}{|c|c|}\\hline\\text{bond enthalpy / kj·mol⁻¹}&\\\\\\hline\\ce{c-h}&413\\\\\\hline\\ce{c-c}&348\\\\\\hline\\ce{c-o}&358\\\\\\hline\\ce{c=o}&799\\\\\\hline\\ce{o-h}&463\\\\\\hline\\ce{o=o}&495\\\\\\hline\\end{array}$$

\\) (a) +1255 kj·mol⁻¹ (b) -1255 kj·mol⁻¹ (c) -1509 kj·mol⁻¹ (d) -2044 kj·mol⁻¹ 5(413) + 358 + 463 + 3(495) - 4(799) + 6(463) =

Explanation:

Step1: Identify Bonds in Reactants

In \( \ce{C2H5OH + 3O2 -> 2CO2 + 3H2O} \), ethanol (\( \ce{C2H5OH} \)) has: 5 \( \ce{C-H} \) (413 kJ/mol), 1 \( \ce{C-C} \) (348 kJ/mol), 1 \( \ce{C-O} \) (358 kJ/mol), 1 \( \ce{O-H} \) (463 kJ/mol). \( \ce{O2} \) has 3 \( \ce{O=O} \) (495 kJ/mol).
Total reactant bond enthalpy:
\( 5\times413 + 348 + 358 + 463 + 3\times495 \)
\( = 2065 + 348 + 358 + 463 + 1485 \)
\( = 2065+348=2413; 2413+358=2771; 2771+463=3234; 3234+1485=4719 \) kJ/mol.

Step2: Identify Bonds in Products

\( \ce{2CO2} \): 2 molecules, each has 2 \( \ce{C=O} \) (799 kJ/mol), so \( 2\times2\times799 = 3196 \) kJ/mol.
\( \ce{3H2O} \): 3 molecules, each has 2 \( \ce{O-H} \) (463 kJ/mol), so \( 3\times2\times463 = 2778 \) kJ/mol.
Total product bond enthalpy: \( 3196 + 2778 = 5974 \) kJ/mol.

Step3: Calculate Enthalpy of Combustion

Enthalpy change (\( \Delta H \)) = Reactant bonds broken - Product bonds formed.
\( \Delta H = 4719 - 5974 = -1255 \) kJ/mol.

Answer:

B. -1255 kJ·mol⁻¹