Test unit 4 for students using book ACE 4 (Oxford)Descripción completa
LDCS theory
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Exercises unit 4 rooftops5. Past simple and jobs.
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This is very useful for people doing Edexcel A2 Chemistry
Descripción: 3º primaria Unit 4 Find out
CS6660 - Artificial Intelligence Anna University - Chennai
ELECTROMAGNETIC THEORY AND WAVES WAVES BEHAVIOR IN GUIDED MEDIUMS AND RADIATION
Version 181 – Unit 4 Exam – vandenbout – (51335) This This prin printt-ou outt shou should ld hav have 20 ques questi tion ons. s. MultipleMultiple-cho choice ice questions questions may contin continue ue on the next column or page – find all choices before answering.
001
4.0 poin points ts
If the change in internal energy for a given process is zero, which of the following must be true?
vandenbout/labrake - ch301
1. q = -w correct
—EXAM4—
2. w = 0
9:30 am class only - 51335
1
3. ∆E = ∆H
7-9pm Dec 5, 2012
4. q = 0 Explanation:
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Thermodynamic Data at 25 C
Substance
◦
∆H f
∆Gf
S
kJ/mol
kJ/mol
J/mol K
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Al (s) — — Al2O3 (s) -1675.7 -1582.4 H2 O (ℓ ( ℓ) -2 8 6 -237 H2 O (g) -242 -229 CO (g) -111 -137 CO2 (g) -394 -395 CH4 (g) -7 5 -51 C3 H8 (g) -104 -24 C6 H12 O2 (ℓ ( ℓ) -5 5 5 — C12 H22O11 (s) -2222 -1545 O2 (g) — — Fe2 O3 (s) -824.2 -742.2 Fe3 O4 (s) -1118.4 – 10 1015.4
Single Bond Energies (kJ/mol of bonds) H C N O S F Cl H 4 36 C 413 346 N 391 305 163 O 463 358 201 146 S 347 22772 — — 226 F 565 485 283 190 284 155 Cl 43 4 32 339 192 218 255 253 242 Multiple Bond Energies (kJ/mol of bonds) C=C 602 C=N 615 C=O 799 C≡C 835 C≡N 887 C≡O 1072 N=N 418 O=O 498 N≡N 945
∆E = q + w = 0 q = -w 002
4.0 poin points ts
Carbon monoxide reacts with oxygen to form carbon dioxide by the following reaction: 2 CO(g) CO(g) + O2 (g) → 2 CO2(g) ∆H for for this reaction is −135 135..28 kcal. If 811.7 kcal is released, how many moles of CO must have reacted? 1. 12.0 mol correct 2. 16.3 mol 3. 6.00 mol 4. 270.5 mol 5. 1.35 mol 6. 40.5 mol 7. 5.41 mol Explanation:
1 mol rxn 811..7kcal) = 12 mol CO ×(−811 so 12 mol CO were reacted. 003
4.0 poin points ts
At constant temperature and pressure, how is ∆S univ ∆Gsys? univ related to ∆G
Version 181 – Unit 4 Exam – vandenbout – (51335) 1. ∆Gsys = ∆S univ
4. -376 kJ/mol
2. ∆Gsys = −T ∆S univ correct
5. -193 kJ/mol Explanation:
3. ∆Gsys = −∆S univ
Cl—F + C≡O −→ Cl—C=O | F Break: Cl—F and C≡O 253 1072
4. ∆Gsys = T ∆S univ
∆S univ T ∆S univ 6. ∆Gsys = T 5. ∆Gsys = −
Make: C=O and C—Cl and C—F 799 339 485
Explanation: 004
2
breaking : +1325 (reactants) making : -1623 (products) 4.0 points
A 22.9 g sample of metal X requires 136 calories of energy to heat it from 10.0 C to 81.0 C. Calculate the specific heat of metal X. 1. 0.2615 2. 0.21 3. 0.2548 4. 0.2038 5. 0.3562 6. 0.3902 7. 0.3081 8. 0.2187 9. 0.35 10. 0.2733
∆H = −298 kJ/mol
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006
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Correct answer: 0.35 J/g · C.
4.0 points
What is the change in entropy (∆S vap ) for the vaporization of ethanol (∆Hvap = 38.6 kJ · mol 1 ) at its standard boiling temperature (78.4 C)? −
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1. −110 J · mol 1 · K 1 −
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2. −492 J · mol 1 · K 1 −
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3. 492 J · mol 1 · K 1 −
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4. 110 J · mol 1 · K 1 correct −
−
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Explanation: 005
Explanation:
∆Hvap T 38, 600 J · mol 1 = = 110 J · mol 351.4 K ∆Svap =
4.0 points
Chlorine monofluoride (ClF) will react with carbon monoxide (CO) to give carbonyl chlorofluoride (COClF):
−
007
Use bond energies (provided elsewhere) to estimate the change in enthalpy (∆H ) for this reaction.
·K
1
−
4.0 points
Given that ∆f H F e O (s) = −826 kJ mol 1 what would you predict for the spontaneity of the following reaction ◦
−
2
ClF + CO −→ COClF
1
−
3
2Fe2 O3 (s) → 4Fe(s) + 3O2 (g) 1. there is no way to predict the spontaneity
1. -298 kJ/mol correct
of this reaction
2. -571 kJ/mol
2. it would be spontaneous only at high temperatures correct
3. -444 kJ/mol
Version 181 – Unit 4 Exam – vandenbout – (51335) 3. it would never be spontaneous
c) 1 mol of pure water
4. it would be spontaneous at all tempera-
1. c < b < a
3
tures 2. a < b < c 5. it would be spontaneous only at low tem-
peratures
3. c < a < b
Explanation:
4. b < c < a
The reaction as written is the reverse of twice the formation of Fe 2O3 . As such this reaction is extremely endothermic (∆H >> 0). However, the entropy change for this reaction should be positive as it generates three moles of gas. The enthalpy changes cause an increase in the free energy, but the entropy change will cause the free energy to decrease. At high temperatures the entropy will dominate and the free energy change will be negative. 008
4.0 points
Which of the following is always true for a spontaneous process at constant temperature? 1. ∆S =
q T
2. ∆S > 0
5. b < a < c 6. a < c < b correct Explanation:
Entropy increases as systems go through endothermic phase transitions and when there is more matter or more dispersed matter present. 010
4.0 points
Consider the reaction Ni(s) + 4 CO(g) → Ni(CO)4(g) Assuming the gases are ideal, calculate the work done on the system at a constant pressure of 1 atm at 75 C for the conversion of 1.00 mole of Ni to Ni(CO) 4. ◦
3. ∆S system + ∆S surroundings > 0 correct
q 4. ∆S < T
q 5. ∆S system + ∆S surroundings = T Explanation:
The entropy of a system or its surroundings may increase or decrease. The Second Law of Thermodynamics states that in spontaneous changes, the universe tends toward a state of greater disorder; that is, entropy increases (∆S > 0).
Calculate the entropy change for the system when 40 g of ethanol are heated from 25 C to 40 C given that the heat capacity for ethanol is 2.44 J g 1 K 1. ◦
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−
−
1. +306 kJ/mol correct
1. 4.8 J K 1 correct
2. +555 kJ/mol
2. -45.9 J K 1
3. +192 kJ/mol
3. 0.12 J K 1
4. +355 kJ/mol
4. 45.9 J K 1
5. +93.5 kJ/mol
5. -2.3 J K 1
6. +413 kJ/mol
Explanation:
7. +592 kJ/mol
−
−
−
−
−
T f The entropy change is ∆S = Cln ∆S = T i
Explanation:
Version 181 – Unit 4 Exam – vandenbout – (51335) O2 (g) → 2 O(g) ∆H = +498.4 kJ/mol NO(g) + O3 (g) → NO2 (g) + O2 (g) ∆H = −200 kJ/mol The standard formation of ozone is ◦
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3 O2 (g) → O3 (g) 2 ∆H = +142.7 kJ/mol We calculate the ∆Hrxn using Hess’ Law: To combine the reactions and get the desired reaction, reverse the second and third equations and add half of the first one: NO2(g) + O2 (g) → NO(g) + O3 (g) ∆H = +200 kJ/mol 3 O3 (g) → O2 (g) 2 ∆H = −142.7 kJ/mol 1 O2 (g) → O(g) 2 1 ∆H = (498.4 kJ/mol) 2 NO2 (g) → NO(g) + O(g) ∆Hrxn = 306.5 kJ/mol 015
4.0 points
What is ∆S univ for a reaction conducted at 100 C and for which ∆S is 170 J · mol 1 · K 1 and ∆H is 43.2 kJ · mol 1. ◦
−
−
016
4.0 points
Which of the following reactions as written corresponds to a standard formation reaction? 1. SO2 (g) +
1 O2 (g) → SO3 (g) 2
2. 2H (g) + O 2 (g) → H2 O2 (ℓ) 3. Na(s) +
1 Cl2 (g) → NaCl(s) correct 2
4. 2 H2 (g) + O2 (g) → 2 H2 O(ℓ) 5. CO2 (s) → CO2 (g) Explanation:
All of the reactants in a formation reaction must be ELEMENTS in their thermochemical standard state. AND, to match table values, only 1 MOLE of product can be formed because standard table values are in kJ/mol. Only the NaCl fits this definition. The H2O reaction unfortunately is forming 2 moles of water instead of one. 017
4.0 points
The enthalpy of fusion of methanol (CH3OH) is 3.16 kJ/mol. How much heat would be absorbed or released upon freezing 25.6 grams of methanol?
−
1. 0.253 kJ absorbed
1. −262 J · mol 1 · K 1
2. 3.95 kJ absorbed
2. 285.8 J · mol 1 · K 1
3. 2.52 kJ absorbed
3. 602 J · mol 1 · K 1
4. 0.253 kJ released
4. 54.2 J · mol 1 · K 1 correct
5. 3.95 kJ released
−
−
−
−
−
−
−
−
Explanation:
6. 2.52 kJ released correct
∆S univ = ∆S sys + ∆S surr 1
Explanation: 1
= 170 J · mol · K 43200 J · mol 1 − 373 K = 54.2 J · mol 1 · K 1 −
−
MW CH3OH = 32.042g
−
−
−
5
∆H(fusion) = −∆H(freezing)
Version 181 – Unit 4 Exam – vandenbout – (51335) q = 25.6 g ×
1mol −3.16 kJ × = −2.525 kJ 32.042g mol 018
4.0 points
A sample of 2 grams of compound Y is burned completely in a bomb calorimeter which contains 2500 g of water. The temperature rises from 25.529 C to 25.991 C. What is ∆U rxn for the combustion of compound Y? The hardware component of the calorimeter has a heat capacity of 3.78 kJ/ C. The specific heat of water is 4.184 J/g · C, and the MW of Y is 129 g/mol. 1. -311.8 2. -572.3 3. -503.5 4. -235.9 5. -424.3 6. -213.5 7. -165.2 8. -552.3 9. -269.5 10. -335.2 ◦
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◦
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Correct answer: −424.3 kJ/mol of Y. Explanation:
SH = 4.184 J/g · C HC = 3.78 kJ/ C mY = 2 g MWY = 129 g/mol mwater = 2500 g ∆T = 25.991 C − 25.529 C = 0.462 C ◦
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Heat = (SH)(m water )(∆T ) + (HC)(∆T ) , Then divide the heat by the number of moles which is mass/MW. 019
4.0 points
Sodium is a metal that reacts violently with water. Chlorine is a toxic, choking gas. So it’s a really good thing that when you put salt on your fries (chips) that it doesnt suddenly turn into sodium metal and chlorine gas. Why is this the case? 1. The standard enthalpy of formation of sodium chloride is negative. 2. The standard Gibbs free energy of formation of sodium chloride is negative. CORRECT
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3. The standard entropy change for the formation of sodium chloride is negative. 4. Salt dissolves in the water in your food. 5. Salt has a really high melting point. Explanation:
The standard Gibbs free energy of formation of sodium chloride is negative so the process is spontaneous. Thus ’unforming’ sodium chloride is nonspontaneous, so sodium chloride is stable.
Version 181 – Unit 4 Exam – vandenbout – (51335) 020
24.0 points
This question is merely a placeholder for the points in the hand-graded portion of the exam. Explanation:
This question is merely a placeholder for the points in the hand-graded portion of the exam.