Chemistry 11 (HL)
Unit 5 / IB Topic 5.4
Energetics 4 – Practice Problems - Answers
Bond Enthalpies
COMPARISON OF EXOTHERMIC AND ENDOTHERMIC REACTIONS
endothermic reaction
exothermic reaction
sign of enthalpy
change (∆H)
positive
negative
enthalpy of
reactants vs
products
Hreactants greater / less than Hproducts
Hreactants greater / less than Hproducts
energy of bond
making vs breaking
energy used to break reactant bonds
greater / less than
energy used to make product bonds
energy used to break reactant bonds
greater / less than
energy used to make product bonds
stability of
reactants
and products*
reactants are more / less stable than
products
reactants are more / less stable than
products
* In general, stability is related to the enthalpy of a substance. The lower the enthalpy of the
substance, the more stable it is.
p. 1
Chemistry 11 (HL)
Unit 5 / IB Topic 5.4
1.
Which of the following is an endothermic process?
+
a) 2 Cl(g) → Cl2(g)
b)
Na(g) → Na (g) + 1 e
+
c) Na (g) + Cl (g) → NaCl(s)
d)
Na(g) → Na(s)
2.
Which equation represents the bond enthalpy for the H-Cl bond?
a) HCl(g) → ½ H2(g) + ½ Cl2(g)
b)
HCl(g) → H(g) + Cl(g)
+
+
+
c) HCl(g) → H (g) + Cl (g)
d)
HCl(aq) → H (aq) + Cl (aq)
3.
Which of the following is equivalent to the bond enthalpy of the carbon-oxygen bond in carbon
monoxide?
a) CO(g) → C(s) + O(g)
b)
CO(g) → C(g) + ½ O2(g)
c) CO(g) → C(s) + ½ O2(g)
d)
CO(g) → C(g) + O(g)
4.
Identify the bonds broken in this reaction:
reactants:
5.
C2H6(g) → 2 C(g) + 6 H(g)
1 x C-C bond; 6 x C-H bond
Use bond enthalpy values to calculate the enthalpy of combustion of these fuels. Write the
balanced equation for the reaction first!
a)
ethane (C2H6):
C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(l)
∆H = ∑BEreactants - ∑BEproducts
= [ 1 (C-C) + 6 (C-H) + 7/2 (O=O) ] – [ 4 (C=O) + 6 (O-H) ]
= [ 347 + 6(413) + 7/2(498) ] – [ 4(746) + 6(464) ]
= 4568 – 5768
= -1200 kJ
b)
methanol (CH3OH):
CH3OH(g) + 3/2 O2(g)  CO2(g) + 2H2O(l)
∆H = ∑BEreactants - ∑BEproducts
= [ 3 (C-H) + (C-O) + (O-H) + 3/2 (O=O) ] – [ 2 (C=O) + 4 (O-H) ]
= [ 3(413) + 358 + 464 + 3/2(498) ] – [ 2(746) + 4(464) ]
= 2808 - 3348
= -540 kJ
p. 2
Chemistry 11 (HL)
6.
Unit 5 / IB Topic 5.4
Compare your answers in 5 to the standard enthalpy of combustion values for ethane and
methanol. (Refer to Table 12 in the Chemistry Data booklet for the standard values.) Suggest
reasons for any difference.
The standard enthalpy of combustion of ethane is -1560 kJ/mol, whereas the value found using
bond enthalpies is lower at -1200 kJ/mol. For methanol, the standard value is -726 kJ/mol vs
-540 kJ/mol using bond enthalpies.
The enthalpy of combustion using bond enthalpy values is LOWER than the standard value.
When making comparisons of ∆H values, only consider the MAGNITUDE of ∆H. The sign only
tells you if the reaction is endothermic or exothermic.
In the actual combustion of ethane, liquid water is formed rather than gaseous water. More
energy would be released because the liquid form of water has a lower KE than the gaseous form.
Therefore the standard value has greater energy released.
The bond enthalpy values are also averages for each bond type, taken from a wide range of
molecules. The value calculated using bond enthalpies will therefore be an estimate.
7.
Use bond enthalpy values to calculate the enthalpy of reaction for
a)
CH4 + Br2 → CH3Br + HBr
∆H = ∑BEreactants - ∑BEproducts
= [ 4 (C-H) + (Br-Br) ] – [ 3 (C-H) + (C-Br) +(H-Br) ]
= [ 4(413) + 193 ] – [ 3(413) + 290 + 366 ]
= 1845 – 1895
= -50 kJ
b)
the formation of hydrogen chloride: ½ H2(g) + ½ Cl2(g)  HCl(g)
∆H = ∑BEreactants - ∑BEproducts
= [ ½ (H-H) + ½ (Cl-Cl) ] – [ (H-Cl) ]
= [ ½ (436) + ½ (243) ] – [ 432 ]
= 339.5 – 432
-1
= -93 kJ mol
8.
Given the reaction:
N2(g) + 3 Cl2(g) → 2 NCl3(g)
∆H = +688 kJ
Calculate the bond enthalpy value for the N-Cl bond.
∆H = ∑BEreactants - ∑BEproducts
688 = [ N-N triple + 3 (Cl-Cl) ] – [ 6 (N-Cl) ]
688 = [ 945 + 3(243) ] – 6x
688 = 1674 – 6x
6x = 986
-1
x = 164 kJ mol
p. 3