Concept explainers
(a)
Interpretation:
For the given reactions, the equation has to be completed and balanced.
Concept Introduction:
Combustion reaction is the one in which, the substance is burnt in presence of oxygen. If any hydrocarbon is burnt in presence of oxygen, the products obtained are carbon dioxide and water only.
(a)
Answer to Problem 23.32QP
The balanced equation for the given reaction can be written as,
Explanation of Solution
The given reaction in the problem statement is an combustion reaction and hence, the product obtained will be carbon dioxide and water only. Hence, the raw equation can be written as,
As the total number of atoms present in both reactants and products must be equal, balancing has to be done. To balance the carbon atom in both sides, the product side has to be multiplied by 4 for carbon dioxide and 5 for water. This leads, balancing of both carbon and hydrogen atoms on both sides.
Oxygen count was thirteen in product side. But in the reactant side it is two. Hence, multiplying by 13/2 on reactant side balances the oxygen atom also on both sides. Therefore, the balanced chemical equation for the given reaction can be given as,
To remove the fraction multiply all the compounds by 2. This gives the balanced equation as,
The given incomplete combustion reaction is completed and the balanced chemical equation also written.
(b)
Interpretation:
For the given reactions, the equation has to be completed and balanced.
Concept Introduction:
Oxidation is loss of electron by a species. This also can be said as addition of oxygen. Reduction is a process in which electrons are gained by a species.
(b)
Answer to Problem 23.32QP
The balanced equation for the given reaction can be written as,
Explanation of Solution
Reduction: MnO4− → MnO2
Balance O: MnO4− + 2H2O → MnO2 + 4OH−
Balance charge: MnO4− + 2H2O + 3e− → MnO2 + 4OH−
Add half-reactions:
Therefore, the complete equation for the given reaction can be given as,
The balanced equation for the given reaction was written.
(c)
Interpretation:
For the given reactions, the equation has to be completed and balanced.
Concept Introduction:
Bromination of
(c)
Answer to Problem 23.32QP
The balanced equation for the given reaction can be written as,
Explanation of Solution
Bromination across the double bond.
The given bromination reaction was completed and the product was written with balanced equation.
(d)
Interpretation:
For the given reactions, the equation has to be completed and balanced.
Concept Introduction:
Nitration of
(d)
Answer to Problem 23.32QP
The balanced equation for the given reaction can be written as,
Explanation of Solution
Nitration of ethylbenzene takes place in presence of nitric acid and sulphuric acid. As ethyl is a para directing group, the nitro group is substituted in the para position. By considering this, the complete reaction can be given as,
The given nitration reaction product was drawn and the equation was completed.
(e)
Interpretation:
For the given reactions, the equation has to be completed and balanced.
Concept Introduction:
Chlorination of aromatic compound happens in presence of catalyst. Usually this does not occur in absence of catalyst. The position of chlorination in the aromatic ring depends upon the group already substituted in the ring.
(e)
Answer to Problem 23.32QP
The balanced equation for the given reaction can be written as,
Explanation of Solution
Chlorination of benzene takes place in presence of catalyst with chlorine. By considering this, the complete reaction can be given as,
The given chlorination reaction was completed and the product was written with balanced equation.
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Chapter 23 Solutions
OWLv2 with Student Solutions Manual eBook for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 4 terms (24 months)
- Please correct answer and don't used hand raitingarrow_forwardPlease correct answer and don't used hand raitingarrow_forward(11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B Bond A Bond C a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest Bond Strongest Bond b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)!arrow_forward
- . 3°C with TH 12. (10pts total) Provide the major product for each reaction depicted below. If no reaction occurs write NR. Assume heat dissipation is carefully controlled in the fluorine reaction. 3H 24 total (30) 24 21 2h • 6H total ● 8H total 34 래 Br2 hv major product will be most Substituted 12 hv Br NR I too weak of a participate in P-1 F₂ hv Statistically most favored product will be major = most subst = thermo favored hydrogen atom abstractor to LL Farrow_forwardFive chemistry project topic that does not involve practicalarrow_forwardPlease correct answer and don't used hand raitingarrow_forward
- Q2. Consider the hydrogenation of ethylene C2H4 + H2 = C2H6 The heats of combustion and molar entropies for the three gases at 298 K are given by: C2H4 C2H6 H2 AH comb/kJ mol¹ -1395 -1550 -243 Sº / J K¹ mol-1 220.7 230.4 131.1 The average heat capacity change, ACP, for the reaction over the temperature range 298-1000 K is 10.9 J K¹ mol¹. Using these data, determine: (a) the standard enthalpy change at 800 K (b) the standard entropy change at 800 K (c) the equilibrium constant at 800 K.arrow_forward13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B Bond A Bond C a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest Bond Strongest Bond b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. c. (5pts) Use principles discussed in lecture, supported by relevant structures, to succinctly explain the why your part b (i) radical is more stable than your part b(ii) radical. Written explanation can be no more than one-two succinct sentence(s)! Googlearrow_forwardPrint Last Name, First Name Initial Statifically more chances to abstract one of these 6H 11. (10pts total) Consider the radical chlorination of 1,3-diethylcyclohexane depicted below. 4 4th total • 6H total 래 • 4H total 21 total ZH 2H Statistical H < 3° C-H weakest - product abstraction here bund leads to thermo favored a) (6pts) How many unique mono-chlorinated products can be formed and what are the structures for the thermodynamically and statistically favored products? Product 6 Number of Unique Mono-Chlorinated Products Thermodynamically Favored Product Statistically Favored Product b) (4pts) Draw the arrow pushing mechanism for the FIRST propagation step (p-1) for the formation of the thermodynamically favored product. Only draw the p-1 step. You do not need to include lone pairs of electrons. No enthalpy calculation necessary H H-Cl Waterfoxarrow_forward
- 10. (5pts) Provide the complete arrow pushing mechanism for the chemical transformation → depicted below Use proper curved arrow notation that explicitly illustrates all bonds being broken, and all bonds formed in the transformation. Also, be sure to include all lone pairs and formal charges on all atoms involved in the flow of electrons. CH3O II HA H CH3O-H H ①arrow_forwardDo the Lone Pairs get added bc its valence e's are a total of 6 for oxygen and that completes it or due to other reasons. How do we know the particular indication of such.arrow_forwardNGLISH b) Identify the bonds present in the molecule drawn (s) above. (break) State the function of the following equipments found in laboratory. Omka) a) Gas mask b) Fire extinguisher c) Safety glasses 4. 60cm³ of oxygen gas diffused through a porous hole in 50 seconds. How long w 80cm³ of sulphur(IV) oxide to diffuse through the same hole under the same conditions (S-32.0.0-16.0) (3 m 5. In an experiment, a piece of magnesium ribbon was cleaned with steel w clean magnesium ribbon was placed in a crucible and completely burnt in oxy cooling the product weighed 4.0g a) Explain why it is necessary to clean magnesium ribbon. Masterclass Holiday assignmen PB 2arrow_forward
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