(COMPLETE) zCHEM
z104
(COMPLETE) CHEM 104 LATEST 2024 z z z z
M0DULE 1,2,3,4,5 AND 6 100% CORRECT z z z z z
(PORTAGE LEARNING) |VERIFIED
z z z
EXAM WITH ANSWERS (STEP BY STEP
z z z z z z
EXPLANATION)
z
Table zof zContents
CHEM z104 zModule z1 ................................................................................................................................................... 1
CHEM z104 zModule z2 ................................................................................................................................................... 8
CHEM z104 zModule z3 ................................................................................................................................................. 13
CHEM z104 zModule z4 ................................................................................................................................................. 18
CHEM z104 zModule z5 ................................................................................................................................................. 25
CHEM z104 zModule z6 ................................................................................................................................................. 29
CHEM z104 zModule z1:
Question z1
In zthe zreaction zof zgaseous zN2O5 zto zyield zNO2 zgas zand zO2 zgas zas zshown zbelow, zthe zfollowing
zdata ztable zis zobtained:
2 zN2O5 → z 4 zNO2 z(g) z z
+ z z O2 z(g)
z(g)
Data zTable
z#2
Time z(sec) [N2O5] [O2]
0 0.300 zM 0
300 0.272 zM 0.014 zM
600 0.224 zM 0.038 zM
900 0.204 zM 0.048 zM
,(COMPLETE) zCHEM
z104
1. Using zthe z[O2] zdata zfrom zthe ztable, zshow zthe zcalculation zof zthe zinstantaneous zrate
zearly zin zthe zreaction z(0 zsecs zto z300 zsec).
2. Using zthe z[O2] zdata zfrom zthe ztable, zshow zthe zcalculation zof zthe zinstantaneous zrate zlate zin zthe
zreaction z(2400 zsecs zto z3000 zsecs).
3. Explain zthe zrelative zvalues zof zthe zearly zinstantaneous zrate zand zthe zlate zinstantaneous zrate.
Your zAnswer:
1. rate z= z(0.014 z- z0) z/ z(300 z- z0) z= z4.67 zx z10-5 zmol/Ls
2. rate z= z(0.090 z- z0.083) z/ z(3000 z- z2400) z= z1.167 zx z10-5 zmol/Ls
3. The zlate zinstantaneous zrate zis zsmaller zthan zthe zearly zinstantaneous zrate.
,(COMPLETE) zCHEM
z104
Question z2
The zfollowing zrate zdata zwas zobtained zfor zthe zhypothetical zreaction: z A z z + z z B z z → z z X z z +
z Y
Experiment z# [A] [B] rate
1 0.50 0.50 2.0
2 1.00 0.50 8.0
3 1.00 1.00 64.0
1. Determine zthe zreaction zorder zwith zrespect zto z[A].
2. Determine zthe zreaction zorder zwith zrespect zto z[B].
3. Write zthe zrate zlaw zin zthe zform zrate z= zk z[A]n z[B]m z(filling zin zthe zcorrect zexponents).
4. Show zthe zcalculation zof zthe zrate zconstant, zk.
Your zAnswer:
rate z= zk z[A]x z [B]y
rate z1 z/ zrate z2 z= zk z[0.50]x z[0.50]y z/ zk z[1.00]x z [0.50]y
2.0 z/ z8.0 z= z[0.50]x z / z[1.00]x
0.25 z= z0.5x
x z= z2
rate z2 z/ zrate z3 z= zk z[1.00]x z[0.50]y z/ zk z[1.00]x z [1.00]y
8.0 z/ z64.0 z= z[0.50]y z/ z[1.00]y
0.125 z= z0.5y
y z= z3
rate z= zk z[A]2 z[B]3
2.0 z= zk z[0.50]2 z[0.50]3
k z= z64
Question z3
ln z[A] z- zln z[A]0 z = z- zk zt 0.693 z= zk zt1/2
An zancient zsample zof zpaper zwas zfound zto zcontain z19.8 z% z14C zcontent zas zcompared zto za
zpresent-day zsample. zThe zt1/2 zfor z
14C zis z5720 zyrs. zShow zthe zcalculation zof zthe zdecay zconstant
z(k) zand zthe zage zof zthe zpaper.
, (COMPLETE) zCHEM
z104
Your zAnswer:
0.693 z= zk zt1/2
0.693 z= zk z(5720)
k z= z1.21 zx z 10-4
ln z[A] z- zln z[A]0 z= z- zk zt
ln z19.8 z- zln z100 z= z- z1.21 zx z10-4 z t
t z= z13, z384 zyears
Question z4
Using zthe zpotential zenergy zdiagram zbelow, zstate zwhether zthe zreaction zdescribed zby zthe zdiagram zis
zendothermic zor zexothermic zand zspontaneous zor znonspontaneous, zbeing zsure zto zexplain zyour zanswer.
Your zAnswer:
The zreaction zis zexothermic zsince zit zhas za znegative zheat zof zreaction zand zit zis znonspontaneous
zbecause zit zhas zrelatively zlarge zEact.
Question z5
Show zthe zcalculation zof zKc zfor zthe zfollowing zreaction zif zan zinitial zreaction zmixture zof z0.800 zmole
zof zCO
and z2.40 zmole zof zH2 zin za z8.00 zliter zcontainer zforms zan zequilibrium zmixture zcontaining z0.309
zmole zof zH2O zand zcorresponding zamounts zof zCO, zH2, zand zCH4.
CO z(g) z z
+ z 3 zH2 z(g) z z
CH4 z(g) z
+ z z H2O z(g)
Your zAnswer:
0.309 zmole zof zH2O zformed z= z0.309 zmole zof zCH4 zformed
z104
(COMPLETE) CHEM 104 LATEST 2024 z z z z
M0DULE 1,2,3,4,5 AND 6 100% CORRECT z z z z z
(PORTAGE LEARNING) |VERIFIED
z z z
EXAM WITH ANSWERS (STEP BY STEP
z z z z z z
EXPLANATION)
z
Table zof zContents
CHEM z104 zModule z1 ................................................................................................................................................... 1
CHEM z104 zModule z2 ................................................................................................................................................... 8
CHEM z104 zModule z3 ................................................................................................................................................. 13
CHEM z104 zModule z4 ................................................................................................................................................. 18
CHEM z104 zModule z5 ................................................................................................................................................. 25
CHEM z104 zModule z6 ................................................................................................................................................. 29
CHEM z104 zModule z1:
Question z1
In zthe zreaction zof zgaseous zN2O5 zto zyield zNO2 zgas zand zO2 zgas zas zshown zbelow, zthe zfollowing
zdata ztable zis zobtained:
2 zN2O5 → z 4 zNO2 z(g) z z
+ z z O2 z(g)
z(g)
Data zTable
z#2
Time z(sec) [N2O5] [O2]
0 0.300 zM 0
300 0.272 zM 0.014 zM
600 0.224 zM 0.038 zM
900 0.204 zM 0.048 zM
,(COMPLETE) zCHEM
z104
1. Using zthe z[O2] zdata zfrom zthe ztable, zshow zthe zcalculation zof zthe zinstantaneous zrate
zearly zin zthe zreaction z(0 zsecs zto z300 zsec).
2. Using zthe z[O2] zdata zfrom zthe ztable, zshow zthe zcalculation zof zthe zinstantaneous zrate zlate zin zthe
zreaction z(2400 zsecs zto z3000 zsecs).
3. Explain zthe zrelative zvalues zof zthe zearly zinstantaneous zrate zand zthe zlate zinstantaneous zrate.
Your zAnswer:
1. rate z= z(0.014 z- z0) z/ z(300 z- z0) z= z4.67 zx z10-5 zmol/Ls
2. rate z= z(0.090 z- z0.083) z/ z(3000 z- z2400) z= z1.167 zx z10-5 zmol/Ls
3. The zlate zinstantaneous zrate zis zsmaller zthan zthe zearly zinstantaneous zrate.
,(COMPLETE) zCHEM
z104
Question z2
The zfollowing zrate zdata zwas zobtained zfor zthe zhypothetical zreaction: z A z z + z z B z z → z z X z z +
z Y
Experiment z# [A] [B] rate
1 0.50 0.50 2.0
2 1.00 0.50 8.0
3 1.00 1.00 64.0
1. Determine zthe zreaction zorder zwith zrespect zto z[A].
2. Determine zthe zreaction zorder zwith zrespect zto z[B].
3. Write zthe zrate zlaw zin zthe zform zrate z= zk z[A]n z[B]m z(filling zin zthe zcorrect zexponents).
4. Show zthe zcalculation zof zthe zrate zconstant, zk.
Your zAnswer:
rate z= zk z[A]x z [B]y
rate z1 z/ zrate z2 z= zk z[0.50]x z[0.50]y z/ zk z[1.00]x z [0.50]y
2.0 z/ z8.0 z= z[0.50]x z / z[1.00]x
0.25 z= z0.5x
x z= z2
rate z2 z/ zrate z3 z= zk z[1.00]x z[0.50]y z/ zk z[1.00]x z [1.00]y
8.0 z/ z64.0 z= z[0.50]y z/ z[1.00]y
0.125 z= z0.5y
y z= z3
rate z= zk z[A]2 z[B]3
2.0 z= zk z[0.50]2 z[0.50]3
k z= z64
Question z3
ln z[A] z- zln z[A]0 z = z- zk zt 0.693 z= zk zt1/2
An zancient zsample zof zpaper zwas zfound zto zcontain z19.8 z% z14C zcontent zas zcompared zto za
zpresent-day zsample. zThe zt1/2 zfor z
14C zis z5720 zyrs. zShow zthe zcalculation zof zthe zdecay zconstant
z(k) zand zthe zage zof zthe zpaper.
, (COMPLETE) zCHEM
z104
Your zAnswer:
0.693 z= zk zt1/2
0.693 z= zk z(5720)
k z= z1.21 zx z 10-4
ln z[A] z- zln z[A]0 z= z- zk zt
ln z19.8 z- zln z100 z= z- z1.21 zx z10-4 z t
t z= z13, z384 zyears
Question z4
Using zthe zpotential zenergy zdiagram zbelow, zstate zwhether zthe zreaction zdescribed zby zthe zdiagram zis
zendothermic zor zexothermic zand zspontaneous zor znonspontaneous, zbeing zsure zto zexplain zyour zanswer.
Your zAnswer:
The zreaction zis zexothermic zsince zit zhas za znegative zheat zof zreaction zand zit zis znonspontaneous
zbecause zit zhas zrelatively zlarge zEact.
Question z5
Show zthe zcalculation zof zKc zfor zthe zfollowing zreaction zif zan zinitial zreaction zmixture zof z0.800 zmole
zof zCO
and z2.40 zmole zof zH2 zin za z8.00 zliter zcontainer zforms zan zequilibrium zmixture zcontaining z0.309
zmole zof zH2O zand zcorresponding zamounts zof zCO, zH2, zand zCH4.
CO z(g) z z
+ z 3 zH2 z(g) z z
CH4 z(g) z
+ z z H2O z(g)
Your zAnswer:
0.309 zmole zof zH2O zformed z= z0.309 zmole zof zCH4 zformed
