ANSWER THE QUESTIONS 1 ∆H of the reaction HCl + NaOH NaCl + H 2 O is calculated based on the molar of HCl or NaOH when 25mL of HCl 2M solution reacts with 25mL of NaOH 1M solution?
Trang 1GENERAL CHEMISTRY LAB REPORT
Instructor: Phạ m Hoàng Huy Phư c Lợi ớ
Class: CC04
Lê Nhật Minh – 1952094
July 23, 2022
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I EXPERIMENTAL RESULTS
1) Experiment 1
Temperature °C First time Second time Third time
m c 0 0 ave = 11.66 cal/K
(Detail calculation of one value of m 0 c 0 )
(m 0 c 0 + mc)(t – t ) = mc(t – t ) 2 3 3 1 ⟺ m 0 c 0 = mc ( t 3 – t 1 ) − t ( 2 – ) t 3
t 2 – t 3 First time: m 0 c 0 = 50
8 ≈ 6,25 끫殠끫殠끫殠/끫歼 Second time: m 0 c 0 = 16.5 50 ≈ 3,03 끫殠끫殠끫殠/끫歼 m 0 c 0 = 6.25 +3 +2 . 3 03 86 ≈ 4.046 끫殠끫殠끫殠/ Third time: m 0 c 0 = 50
17.5 ≈ 2,86 끫殠끫殠끫殠/끫歼
2) Experiment 2
Temperature °C First time Second time Third time
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If t 1 t ≠ 2 then ∆t is calculated as the difference between t 0 and
끫欪 끫殞 +끫欪 끫殠
끫 殠
(Detail calculation of one value of Q)
Q = (m 0 c 0 + mc)(t 3 − t 1 + t 2 2 )
∆H = −Q n First time:
Q 1 = (m 0 c 0 + mc) �t 3 − t 1 +t 2
2 � = (4.046 + 50.1,02) �30 − 29+29 2 � = 55.046 끫殠끫殠끫
∆H 1 = −Q 1
n =
−55.046 0,025 = −2201.84 끫殠끫殠끫殠/끫殴끫殴끫殠 Second time:
Q 2 = (m 0 c 0 + mc) �t 3 − t 1 +t 2
2 � = (4.046 + 50.1,02)(34 − 28.5) = 302.753 끫殠끫殠끫
∆H 2 = −Q 2
n =
−302.753 0,025 = −12110.12 끫殠끫殠끫殠 끫殴끫殴끫殠 / Third time:
Q 3 = (m 0 c 0 + mc) �t 3 − t 1 +t 2
2 � = (4.046 + 50.1,02)(34 − 28.75) = 288,991 끫殠끫殠
∆H 3 = −Q n 3 = −313,3
0,025 = −11559.66 끫殠끫殠끫殠/끫殴끫殴끫殠
⟹ Q = 215.597 끫殠끫殠끫殠
⟹ ∆H = −8623,873 끫殠끫殠끫殠/끫殴끫殴끫殠
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3) Experiment 3
Temperature °C First time Second time Third time
∆H (cal/mol) -18384.8 -15758.4 -13132.0
(Detail calculation of one value of Q and ∆H)
끫 欠 = �끫歌 끫殜 끫欸 끫 殜 + 끫歌 끫欎 끫殠 끫欜 끫欸 끫欎 끫殠 끫 欜 + 끫歌 끫欄끫欄끫
First time: Q 1 = �m 0 c 0 + m H 2 O H c 2 O + m CuSO 4 c CuSO 4 �(t 2 − t 1 )
= (4.046 + 50.1 + 3.97)(35 − 30) = 290.58 끫殠끫殠끫殠
∆끫歶 1 = −끫殈 1
끫 殶 =
−290.58 0,025 = −11623.2 끫殠끫殠끫殠/끫殴끫殴끫殠 Second time: Q 2 = �m 0 c 0 + m H 2 O H c 2 O + m CuSO 4 c CuSO 4 �(t 2 − t 1 )
= (4.046 + 50.1 + 4.05)(35 − 30) = 290.98 끫殠끫殠끫殠
∆끫歶 2 = −끫殈 2
끫 殶 = −290.98 0,025 = −11639.2 끫殠끫殠끫殠 끫殴끫殴끫殠 / Third time: Q 3 = �m 0 c 0 + m H 2 O H c 2 O + m CuSO 4 c CuSO 4 �(t 2 − t 1 )
= (4.046 + 50.1 + 4.03)(35 − 30) = 290.88 끫殠끫殠끫殠
∆끫歶 3 = −끫殈 3
끫 殶 = −290.88 0,025 = −11635.2 끫殠끫殠끫殠/끫殴끫殴끫殠
⟹ ∆H = − 11623. 2+− 11639. 3 2+− 11635.2 = −11632.54 cal/mol
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4) Experiment 4
Temperature °C First time Second time Third time
(Detail calculation of one value of Q and ∆H)
끫 欠 = �끫歌 끫殜 끫欸 끫 殜 + 끫歌 끫欎 끫殠 끫欜 끫欸 끫欎 끫殠 끫 欜 + 끫歌 끫欚끫欎 끫 First time: Q 1 = �m 0 c 0 + m H 2 O H c 2 O + m NH 4 끫歬끫歬 끫殂끫殂 c 4 끫歬끫歬 �(t 2 − t 1 )
= (4.046 + 50.1 + 4.01)(25.5 − 30) = − 261.702 끫殠끫殠끫殠
∆H 1 = −Q 1
n =
− − ( 261.702) 0,075 ≈ 3489.36 끫殠끫殠끫殠/끫殴끫殴끫殠 Second time: Q 2 = �m 0 c 0 + m H 2 O c H 2 O + m NH 4 끫歬끫歬 c 끫殂끫殂 4 끫歬끫歬 �(t 2 − t 1 )
= (4.046 + 50.1 + 3.98)(25 − 30) = − 290.63 끫殠끫殠끫殠
∆H 2 = −Q 2
n =
−( −290 63) 0,075 ≈ 3875.06 끫殠끫殠끫殠/끫殴끫殴끫殠 Third time: Q 3 = �m 0 c 0 + m H 2 O H c 2 O + m NH 4 끫歬끫歬 끫殂끫殂 c 4 끫歬끫歬 �(t 2 − t 1 )
= (4.046 + 50.1 + 4.03)(25.5 − 30) = − 261.792 끫殠끫殠끫殠
∆H 3 = −Q 3
n = −(
)
−261 792 0,075 ≈ 3490.56 끫殠끫殠끫殠/끫殴끫殴끫殠
⟹ ∆H = 3489 36 3875 + 06 +3490 56
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II ANSWER THE QUESTIONS 1) ∆H of the reaction HCl + NaOH NaCl + H 2 O is calculated based on the molar of HCl
or NaOH when 25mL of HCl 2M solution reacts with 25mL of NaOH 1M solution? Explain your answer
(rest) 0.025 0
∆H of the reaction is calculated based on the molar of NaOH because there is too much HCl
so we cannot calculate ∆H base on the molar of HCl
2) If HCl 1M is replaced by HNO 3 1M, the result of experiment 2 will change or not?
It will not change because HCl and HNO 3 are all strong acids which dissociate completely
in water at moderate concentrations And the reaction between NaOH and HNO 3 is the neutralization reaction
3) Calculate ∆H 3 based on Hess’s law Compare the calculated value to the experimental results Considering six factors that might cause the error
- Heat loss due to the calorimeter
- Thermometer
- Volumetric glassware
- Balance
- Copper (II) sulfate absorbs water
- Assume specific heat of copper (II) sulfate is 1 cal/mol.K
In your opinion, which one is the most significant? Explain your answer? Are there any other factors?
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Hess’s law: ∆H 3 = ∆ H 1 + ∆ H 2 = −18,7 + 2,8 = −15,9 끫殰끫殠끫殠끫殠/끫殴끫殴끫殠 Experimental result: ∆H 3 = −13,132 끫殰끫殠끫殠끫殠
From our view, Copper (II) sulfate absorbs water is the most important reason because in the room temperature the humidity is quite high so the CuSO4 which we used is dried, so it absorbs water from the surroundings when it contacts with the air, it also releases a big amount
of heat This will impact on our result when we do the experiment
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Experimental report of unit 4
Date: 23/07/2022 Class: CC04
I EXPERIMENTAL RESULTS
1) Reaction order with respect to Na 2 S 2 O 3
No Initial concentration (M) ∆ t 1 ∆t 2 ∆ t 3 ∆t ave
Na 2 S 2 O 3 H 2 SO 4
1 0.0125 0.4 24.95s 25.90s 25.10s 25.317s
2 0.025 0.4 11.90s 12.10s 12.20s 12.067s
From ∆t ave of experiment 1 and 2, determine m 1 (a sample of calculation)
끫殴 1 =
lg �∆ 끫毂 끫殜끫殜끫殜1
∆끫毂 끫殜끫殜끫殜2 �
lg � 25.317 12.067�
끫殠끫殲2 ≈ 1.069 From ∆t ave of experiment 2 and 3, determine m 2
끫殴 2 =
lg (∆ 끫毂 끫殜끫殜끫殜2
∆끫毂 끫殜끫殜끫殜3 )
lg ( 12.067 8.967 ) 끫殠끫殲2 ≈ 0.428 Reaction order with respect to Na 2 S 2 O 3 = 끫殴 +끫殴 1 2
2 =
1.069 +0.428
2 ≈ 0.7485
2) Reaction order with respect to H 2 SO 4
No Initial concentration (M) ∆ t 1 ∆t 2 ∆ t 3 ∆t ave
Na 2 S 2 O 3 H 2 SO 4
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From ∆t ave of experiment 1 and 2, determine n 1 (sample calculation)
끫殶 1 =
lg �∆ 끫毂 끫殜끫殜끫殜1
∆끫毂 끫殜끫殜끫殜2 �
lg � 56.247 34.790�
끫殠끫殲2 ≈ 0.693 From ∆t ave of experiment 2 and 3, determine n 2
끫殶 2 =
lg �∆ 끫毂 끫殜끫殜끫殜1
∆끫毂 끫殜끫殜끫殜2 �
lg � 34.790 27.303�
끫殠끫殲2 ≈ 0.350 Reaction order with respect to H 2 SO 4 = 끫殶 +끫殶 1 2
2 = 0.
693+0 350
2 ≈ 0.5215
II ANSWER THE QUESTION 1) In the experiment above, what is the effect of the concentrations of Na 2 S 2 O 3 and H 2 SO 4
on the reaction rate? Write the reaction rate expression Determine the orders of the reaction
The concentration of Na 2 S 2 O 3 is directly proportional to the reaction rate However, the concentration of H 2 SO 4 does not affect the reaction rate Reaction rate expression:
끫 殒 = 끫殰[끫殂끫殠 2 끫殌 2 끫殄 3 ] 0.9945 ∗ [끫歶 2 끫殌끫殄 4 ] 0.112
The order of the reaction is 0.7485 + 0.5215 = 1.2700
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2) Mechanism of the reaction can be written as
H 2 SO 4 + Na 2 S 2 O 3 Na 2 SO 4 + H 2 S 2 O 3
H 2 S 2 O 3 H 2 SO 3 + S
Based on the experimental results, may we conclude that the reaction (1) and (2) is the rate-determining step, which is the slowest step of the reaction? Recall that in the experiments, the amount of the acid H 2 SO 4 is always used in excess.
Reaction (1) is the ion exchange reaction, so its rate is fast Reaction (2) is self oxidation reduction reaction so its rate is slow –
So, reaction (2) decided the rate and also it is the lowest step of the reaction
3) Based on the principle of the experimental method, the reaction rate is considered as
an instantaneous rate or average rate
The reaction rate is calculated by ∆끫殠
∆끫毂 with ∆끫殠 ≈ 0 so the reaction is considered a instantaneous rate
4) If the order of adding H 2 SO and Na S O is reversed, does the reaction order change? 4 2 2 3 Explain your answer
• The reaction order will not change
• In the determined temperature, the relation only depends on the nature of the system (concentration, temperature, surface area, pressure, catalysis), and does not depend on the order
of reactions
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Experimental report of unit 8
Date: 22/07/2002 Class: CC04
I EXPERIMENTAL RESULTS
2) Experiment 2
No V HCl (mL) V NaOH (mL) C NaOH (N) C HCl (N) Deviation
1 st : 끫歬 끫殂끫歬끫歬 끫殒 = 끫殒 끫殂끫殂끫殂끫殂 끫歬 끫殂끫殂끫殂끫殂
끫殂끫歶끫歶 = 10.6 ∗0 1
10 = 0.106 (끫殂)
2 nd : 끫歬 끫殂끫歬끫歬 끫殒 끫殂끫歶끫歶 = 끫殒 끫殂끫殂끫殂끫殂 = 10.7 ∗0 1 끫歬 끫殂끫殂끫殂끫殂
10 = 0.107 ( 끫殂)
3) Experiment 3
No V HCl (mL) V NaOH (mL) C NaOH (N) C HCl (N) Deviation
1 st : 끫歬 끫殂끫歬끫歬 끫殒 끫殂끫歶끫歶 = 끫殒 끫殂끫殂끫殂끫殂 = 10.3 ∗0 1 끫歬 끫殂끫殂끫殂끫殂
10 = 0.103 ( 끫殂)
2 nd : 끫歬 끫殂끫歬끫歬 끫殒 = 끫殒 끫殂끫殂끫殂끫殂 끫歬 끫殂끫殂끫殂끫殂
끫殂끫歶끫歶 = 10.2 ∗0 1
10 = 0.102 (끫殂)
4) Experiment 4
No Indicator V CH3COOH (mL) V NaOH (N) C HCl (N) C CH3COOH (N)
1 st : 끫歬 끫歬끫殂 3 끫歬끫歬끫歬끫殂 끫殒 = 끫殒 끫殂끫殂끫殂끫殂 끫歬 끫殂끫殂끫殂끫殂
10.6 ∗0 1
10 = 0.106 (끫殂)
2 nd : 끫歬 끫歬끫殂 3 끫歬끫歬끫歬끫殂 끫殒 = 끫殒 끫殂끫殂끫殂끫殂 끫歬 끫殂끫殂끫殂끫殂
2 7 ∗0 1
10 = 0.027 (끫殂)
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II ANSWER THE QUESTIONS 1) When changing the concentration of HCl and NaOH, does the titration curve change? Explain
If we change the concentration of HCl or NaOH, the titration curve will not change The titration reaction is 끫歶끫歬끫殠 + 끫殂끫殠끫殄끫歶 → 끫殂끫殠끫歬끫殠 + 끫歶 2 끫殄
끫歬 끫殂끫歬끫歬 끫殒 끫殂끫歬끫歬 = 끫歬 끫殂끫殜끫歬끫殂 끫殒 끫殂끫殜끫歬끫殂 Because 끫殒 끫殂끫歬끫歬 and 끫歬 끫殂끫歬끫歬 are constant So, if 끫歬 끫殂끫歬끫歬 increased, 끫殒
is also true
Therefore, we can conclude that only the pH jump changed and the titration curve still remain
2) The determination of the concentration of HCl in experiments 2 and 3, which one is more precise?
Phenolphthalein is more precise than orange methyl because of 2 reasons:
• The pH jump of phenolphthalein is at about 8 to 10, while that of orange methyl is at about 3.1 to 4.4 although its equivalence point is above 7 (as weal acid reacts with strong base)
• Phenolphthalein easily helps us determine the change by their obvious color Hence,
it gives us more accurate results
Therefore, the determination of concentration of HCl in experiment 2 is more precise than that of experiment 3