Question #8dedd

1 Answer
Jun 4, 2017

Answer:

Warning! Long Answer. We must break the strong #"C-O"# and #"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"#
#"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"#
#BE"/kJ·mol"^"-1":color(white)(m)413color(white)(mmll)347color(white)(mmll)358color(white)(mmll)467color(white)(mmll)495color(white)(mmml)799color(white)(mml)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)|)))" "#

#sumBE_text(reactants) = "[(5×413) + (1×347) + (1×358) + (1×467) + (3× 495)] kJ" = "(2065 + 347 + 358 + 467 + 1485) kJ" = "4722 kJ"#

#sumBE_text(products) = "[(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

#color(white)(mmmmmmmmll)"2CH"_3"CH"_3color(white)(mm) +color(white)(ll) "7O"_2→ color(white)(m)"4CO"_2 + color(white)(l)"6H"_2"O"#
#"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"#
#BE"/kJ·mol"^"-1":color(white)(mm)413color(white)(mmll)347color(white)(mmll)495color(white)(mmml)799color(white)(mmm)467#

#sumBE_text(reactants) = "[(12×413) + (2×347) + (7×495)] kJ" = "(4956 + 694 + 3465) kJ" = "9115 kJ"#

#sumBE_text(products) = "[(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 #"2 mol CO"_2 + "3 mol H"_2"O"#.

The energy difference must come from the bonds broken.

We must break the strong #"C-O"# and #"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")#.