(a)
Interpretation:
The rate constant, half life and the concentration of
Concept Introduction:
The
The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of
Integrated rate law for second order reactions:
Taking in the example of following reaction,
And the reaction follows second order rate law,
Then the relationship between the concentration of
The above expression is called as integrated rate for second order reactions.
Half life for second order reactions:
In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).
The half-life of second order reaction can be calculated using the equation,
Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.
To calculate the rate constant of the reaction
(a)
Answer to Problem 13.146QP
Answer
The rate constant of the reaction is
Explanation of Solution
The plot of
(b)
Interpretation:
The rate constant, half life and the concentration of
Concept Introduction:
The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time. The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.
The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of
Integrated rate law for second order reactions:
Taking in the example of following reaction,
And the reaction follows second order rate law,
Then the relationship between the concentration of
The above expression is called as integrated rate for second order reactions.
Half life for second order reactions:
In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).
The half-life of second order reaction can be calculated using the equation,
Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.
To calculate the half life of the reaction
(b)
Answer to Problem 13.146QP
Answer
The half period of the reaction is
Explanation of Solution
Initial concentration =
Rate constant=
The half-life of second order reaction can be calculated using the equation,
The half period of the reaction =
(c)
Interpretation:
The rate constant, half life and the concentration of
Concept Introduction:
The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time. The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.
The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of
Integrated rate law for second order reactions:
Taking in the example of following reaction,
And the reaction follows second order rate law,
Then the relationship between the concentration of
The above expression is called as integrated rate for second order reactions.
Half life for second order reactions:
In second order reaction, the half-life is inversely proportional to the initial concentration of the reactant (A).
The half-life of second order reaction can be calculated using the equation,
Since the reactant will be consumed in lesser amount of time, these reactions will have shorter half-life.
To calculate the concentration of
(c)
Answer to Problem 13.146QP
Answer
The concentration of
Explanation of Solution
The equation for second order reaction is given as,
Initial concentration =
Rate constant=
The concentration of
The concentration of
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Chapter 13 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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