ENTHALPY CHANGES 3
1)
Find Hf of butane given that the following data.
Hc: C4H10(g) = –2877, C(s) = –394, H2(g) = –286 kJ mol-1
Hess’s Law Calculations (most questions fit one of the following types)
2)
a) Finding H for a reaction when Hf of all reactants and products is known
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
H = [Sum of Hf products] – [Sum Hf reactants]
Hf: C3H8(g) = –104, CO2(g) = –394, H2O(l) = –286 kJ mol-1
Remember that Hf of all elements is zero. Also, watch for the very frequent mistake
of doing reactants – products, rather than products – reactants.
b) Questions involving Hc 
H
reactants
The reaction involved across the top is often
a Hf. You must be very careful to ensure
your arrows go the right way. Remember
that the sum of the clockwise arrows equals
the sum of the anticlockwise arrows.
Find H for the following reaction using the data below.
Hc
3)
Calculate Hf of CCl4(l) given the following data.
CCl4(l) → CCl4(g) = +31 kJ mol-1
C(s) → C(g) = +715 kJ mol-1
Bond enthalpy (Cl-Cl) = +242 kJ mol-1
Bond enthalpy (C-Cl) = +338 kJ mol-1
products
Hc
oxides
4)
Find Hc of propan-2-ol given that the following data.
Hf: CH3CH(OH)CH3(l) = –318, Hc: C(s) = –394, H2(g) = –286 kJ mol-1
c) Questions involving bond enthalpies
This cycle works for any question that
involves bond enthalpies, whether to find a
bond enthalpy or H for a reaction.
Remember that substances must be in the
gas state before bonds are broken, and so
H to go to the gas state is needed for solids
and liquids.
H
reactants
products
5)
C(s) → C(g) = +715 kJ mol-1
Bond enthalpy: (C-H) = +412, (N-H) = +388, (H-H) = +436, (N≡N) = 944 kJ mol-1
Hf CH3NH2(g) = –23 kJ mol-1
(H→gas) +
bond enthalpies
(H→gas) +
bond enthalpies
Calculate the bond enthalpy of the C-N bond in CH3NH2(g) given the following data.
gas atoms
6)
Find H for the hydrogenation of propene using the data below.
CH3CH=CH2(g) + H2(g) → CH3CH2CH3(g)
Calorimetry questions
Hc: CH3CH=CH2(g) = –2059, H2(g) = –286, CH3CH2CH3(g) = –2220 kJ mol-1
To find the heat given out (or taken in) in a reaction,
use this equation. Be careful with m, which is the
mass of water heated (or cooled). Sometimes you
may be given heat capacity (which equals mc)
rather than specific heat capacity (which is c).
When finding H (i.e. heat given out or taken in by
one mole), be careful over units (check if q is in J or
kJ), and remember H is –ve if heat is given out.
q =
mcT
q
m
c
T
heat given out (or taken in)
mass of water heated
specific heat capacity
temperature change
7)
0.55 g of propanone was burned in a calorimeter containing 80 g of water. The
temperature rose by 47.3ºC. Calculate Hc for propanone given the specific heat
capacity of water is 4.18 J mol-1 K-1.
q / moles
8)
25 cm3 of 2.0 mol dm-3 nitric acid was reacted with 25 cm 3 of 2.0 mol dm-3 potassium
hydroxide is an insulated cup. The temperature rose from 20.2ºC to 33.9ºC.
Calculate H for the reaction given the specific heat capacity of water is 4.18 J mol-1
K-1.
=
=
=
=
H =
1)
Hf
4 C(s) + 5 H2(g)
5)
C4H10(g)
4 CO2(g) + 5 H2O(l)
Hf – 2877 = 4(–394) + 5(–286)
Hf
–23
CH3NH2(g)
C-N
3(412)
2(388)
715
2½(436)
½(944)
–2877
4(–394)
5(–286)


C(s) + 2½ H2(g) + ½ N2(g)
C(g) + 5 H(g) + N(g)

= 4(–394) + 5(–286) + 2877


= –129 kJ mol-1

–23 + (C-N) + 3(412) + 2 (388) = 715 + 2½(436) + ½(944)
(C-N)
= 715 + 2½(436) + ½(944) + 23 – 3(412) – 2(388)
= 288 kJ mol-1
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
2)


6)
H = [Sum of Hf products] – [Sum Hf reactants]
CH3CH=CH2(g) + H2(g)
= [ 3(-394) + 4(-286) ] – [–104]
–2220
3 CO2(g) + 4 H2O(l)
Hf
C(s) + 2 Cl2(g)
CH3CH2CH3(g)
–2059
–286
= –2222 kJ mol-1
3)
H
CCl4(l)


H – 2220 = –2059 + –286
+31
4(338)
715
2(242)
H
= –2059 – 286 + 2220
= –125 kJ mol-1
C(g) + 4 Cl(g)


7)
Hf + 31 + 4(338) = 715 + 2(242)
Hf
q
= 715 + 2(242) – 31 – 4(338)
= –184 kJ mol-1
=
mcT
=
80 x 4.18 x 47.3 = 15820 J = 15.82 kJ
moles CH3COCH3 = mass / Mr = 0.55 / 58.0 = 0.00948
H =
4)
3 C(s) + 4 H2(g) + ½ O2(g)
–318
CH3CH(OH)CH3(l)
Hc
3(–394)
4(–286)
8)
3 CO2(g) + 4 H2O(l)


– 318 + Hc = 3(-394) + 4(-286)
Hc

= 3(–394) + 4(–286) + 318
= –2008 kJ
mol-1
q
q / moles = –15.82 / 0.00948
=
– 1669 kJ mol-1
=
mcT
=
50 x 4.18 x (33.9 – 20.2) = 2863 J = 2.863 kJ
moles KOH = conc x vol (dm3) = 2.0 x 25/1000 = 0.0500
moles HNO3 = conc x vol (dm3) = 2.0 x 25/1000 = 0.0500
H =
=
q / moles = –2.863 / 0.0500
– 57.3 kJ mol-1