# Question 8dedd

Jun 4, 2017

Warning! Long Answer. We must break the strong $\text{C-O}$ and $\text{O-H}$ bonds in ethanol that are not present in ethane, so
Δ_text(c)H("ethanol") is less negative than Δ_text(c)H("ethane").

#### Explanation:

Let's calculate Δ_text(c)H for each reaction using the bond energies.

A. Δ_text(c)H for ethanol

color(white)(mmmmmmmmmll)"CH"_3"CH"_2"OH" +color(white)(mmmmmmmml) "3O"_2→ color(white)(m l)"2CO"_2 + "3H"_2"O"
$\text{Bonds:"color(white)(mmmm) "5 C-H" + "1 C-C" + "1 C-O" + "1 O-H" color(white)(m)"3 O=O"color(white)(mll)"4 C=O"color(white)(m)"6 O-H}$
$B E \text{/kJ·mol"^"-1} : \textcolor{w h i t e}{m} 413 \textcolor{w h i t e}{m m l l} 347 \textcolor{w h i t e}{m m l l} 358 \textcolor{w h i t e}{m m l l} 467 \textcolor{w h i t e}{m m l l} 495 \textcolor{w h i t e}{m m m l} 799 \textcolor{w h i t e}{m m l} 467$

The formula for Δ_text(c)H is

color(blue)(bar(ul(|color(white)(a/a)Δ_text(c)H = sumBE_text(reactants) - sumBE_text(products)color(white)(a/a)|)))" "

$\sum B {E}_{\textrm{r e a c \tan t s}} = \text{[(5×413) + (1×347) + (1×358) + (1×467) + (3× 495)] kJ" = "(2065 + 347 + 358 + 467 + 1485) kJ" = "4722 kJ}$

$\sum B {E}_{\textrm{\prod u c t s}} = \text{[(4×799) + (6×467)] kJ" = "(3196 + 2802) kJ" = "5998 kJ}$

Δ_text(c)H = "4722 kJ - 5998 kJ = -1276 kJ/mol"

B. Δ_text(c)H for ethane

$\textcolor{w h i t e}{m m m m m m m m l l} \text{2CH"_3"CH"_3color(white)(mm) +color(white)(ll) "7O"_2→ color(white)(m)"4CO"_2 + color(white)(l)"6H"_2"O}$
$\text{Bonds:"color(white)(mmmmll) "12 C-H" + "2 C-C" color(white)(m)"7 O=O"color(white)(mll)"8 C=O"color(white)(m)"12 O-H}$
$B E \text{/kJ·mol"^"-1} : \textcolor{w h i t e}{m m} 413 \textcolor{w h i t e}{m m l l} 347 \textcolor{w h i t e}{m m l l} 495 \textcolor{w h i t e}{m m m l} 799 \textcolor{w h i t e}{m m m} 467$

$\sum B {E}_{\textrm{r e a c \tan t s}} = \text{[(12×413) + (2×347) + (7×495)] kJ" = "(4956 + 694 + 3465) kJ" = "9115 kJ}$

$\sum B {E}_{\textrm{\prod u c t s}} = \text{[(8×799) + (12×467)] kJ" = "(6392 + 5604) kJ" = "11 996 kJ}$

Δ_text(c)H = "9115 kJ - 11 996 kJ = -2881 kJ"

This is the value of Δ_text(c)H for 2 mol of ethane.

Δ_text(c)H for ethane is -1440 kJ/mol.

Why the difference?

Δ_text(c)H("ethanol") - Δ_text(c)H("ethane") = "(-1276 – 1440) kJ/mol = 168 kJ/mol"

1 mol of ethanol and 1 mol of ethane each form $\text{2 mol CO"_2 + "3 mol H"_2"O}$.

The energy difference must come from the bonds broken.

We must break the strong $\text{C-O}$ and $\text{O-H}$ bonds in ethanol that are not present in ethane, so

Δ_text(c)H("ethanol") is less negative than Δ_text(c)H("ethane")#.