Chapter 15: Hess's Law and
Standard Enthalpies of
Formation Worksheet KEY
There are 3 mistakes on this
key. If you find any of them
and show me I will give you a
piece of candy.
/
'tfoY
NAME
DATE __ ~
~
_
PERIOD
c: ?f-S)
_
PoSS
_
,6/--tZ
CHAPTER 15: HESS'S LAW AND STANDARD ENTHAlPIES OF FORMATION
Answer the questions below. Remember that you must show work to get points. Don't forget units!
1. Use the equations below to determine ~H for the reaction: 2CO(g) + 2NO(g) ~ 2COz(g) + Nz(g)
= -566.0
a. 2CO(g) + Oz(g) ~ 2COz(g) ~H
b. Nz(g) + Oz(g) ~ 2NO(g) ~H
2CO(,cl
. 0l
_ 2 ;\JO~
_
).i23)
2)1°0J+ZCO«)
AI-I:::.~S-6b"O e.r:
+ Oz 0)
AH
~
2L02(;
~
kJ
2(OlU)
2 /lPrJj-t 2(O~)+ ~
2. Use the equations
= -180.6
kJ
c::
;[50.6
2(O2~)
)J2~)+-~.Jl> ..f-
N
L\J-}-= -3\S~'Io
A/-z-I8)
H for the reaction: C (5, diamond) ~ C (s, graphite)
elow to determine
= -394 kJ
~H = -396 kJ
a. C (s, graphite) + Oz(g) ~ COz(g) ~H
b. C (s, diamond) + Oz(g) ~ COz(g)
AI-J=- "591/ferc{s)J;Qmof1c() + 02 fJJ ~
C02{gJ ~ If=- -396 /t--SCc¥!J 1- (({;I j/Q;noN/)+ :J4S 4- C Cr'(jro/'Me) r-~..f fo/j) .6Ji
C (sj diamond) -7 C (t-J8rqfi/~
.
C(~/O(Qf)',IeJ +- OzrgJ
COzIJ) ~
-Z.ol&
.
3. ~H for the reaction: 4AI(s) + 3MnOz(s) ~ 2Alz03(s) + 3Mn(s) is -1789 kJ. Use ~H from this reaction
to determine ~H for the second equation below.
a. 4AI(s) + 30z(g)
~ 2A!z03(S) ~H = -3352 kJ
b. Mn(s) + Oz(g) ~ MnOz(s)
4AJ
-r 2°1-
sjVIn 02-
+- Lj/rl
4 AI + 3Mn
+-
-/1C611t: J; - 5SS-2
. -I--ssD~T-I-3
_~~
50 -:? ~J-;;
2~
I
)
-?1S-2 f:J
,,6 If;; J (;<) t J ,6 fJ-:;.
sj'{
O2
L1 tt-; -S--2/ ~ J
A:;
5
?AII :}-'50 z
3 Mtl0z
)
/1,
--7 Z
~H =?
?AJ2-{)~ ~ sfi1Y] + ~
?~JZ
Os ..f- s)A~
+- 2. X
/60
3 (t;ZI /t--r)
-=-1,5(p?,}£.j
AN-:: -I 7~1J..:T
S fo:.T ~ U x=.bH::
S'
_~ZliOD~
4.
Using the standard enthalpies of formation
below calculate 6H\xn for the reaction:
4NH3(g) + 702(g) -7 4N02(g) + 6H20(1)
= -45.9
a. 6WdNH3)
kJ/mol
= 0.0 kJ/mol
(N02) = 33.2 kl/rnol
(H20) = -285.8 kJ/mol
b. 6Wf (02)
c. 6Wf
\
d. 6Wf
;\ #~" = [Mbill
Att
J- [[V) ~II/
(AlO,Jr (r;)~lIfo(If 0fl
(AlII').;- h).J ~; (o.)
_
rl{fO -
_
frlf)( '15: 9 fr)+ (7) (60 JJJ] - [('/)(n.
DJI~xlI ~ - ) <ts. 0 is: + 0
2 kJ) -tfG){-
2?,5:g-J<.J)]
/3'2. l?' !::-J + 17 It;, 8' f:.-:r
1"3CJg. '-10 es:
J ~Xh:/
5
/L'J -
~ '..J
Using the standard enthalpies of formation
. [4C(s) + 4H2(g) + 02(g) ~ C3H7COOH(I) 6H
below and the following equation:
= -534
.
.
kl] calculate 6Wcomb for butanoic acid
(C3H7COOH(I)+ 502(g) ~ 4C02(g) + 4H20(1)) .
a. 6Wf (C) = 0.0 kJ/mol
b. 6Wf (H2) = 0.0 kl/rnol
c. 6Wf (02)
= 0.0
kJ/mol
d. 6Wf (C3H7COOH)= ?
e. 6Wf (C02)
f.
I
6~;h
-e
= -393.5
kJ/mol
6Wf (H20) = -285.8 kJ/mol
[CI)!J1f$ (G?/!-7{()Of./)] ~[(If)1:,!f°
(c.).d'l)Mp'(J.I.,) I {fM!Ii {oJ]
[J (O.6NJ +- Lj(().O~JJ+-c. O~T]
-t;?/-l IX ;-X ::..-M e (C~1/ Goo H)
f
~;:;$t;f--J ~
X-
7
LJ))!-b z [rtfJAJ; '(UJ) -f [l/jA/If' (I{ 0J]-[fl JJjl/ (~lIloolf)i. (S~/(P (Ii)J
6
111+;0•.b ~Ifl/j{- 39?-:;ff);(lJ)(- Z8S,UJ)] + ~5t.j JJ - 0,6 ~ J
D!I;()~b -=- - 1:;-7t! tI - }1'/], ? k-J-I- 554l J -()~O k J
L If;o~£ - - 2/ 'l3 2D k-J
J
Z P+St