EBK INTRODUCTORY CHEMISTRY
8th Edition
ISBN: 9780134553153
Author: CORWIN
Publisher: PEARSON CO
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Chapter 12, Problem 18ST
Interpretation Introduction
Interpretation:
The molecular shape of a water molecule,
Concept introduction:
The VSEPR theory talks about the shape of the molecule obtained due to the repulsion between the electron pairs surrounding the central atom. Electron pair geometry is the shape obtained after the arrangement bonding and nonbonding electron pairs. Molecular geometry is the shape obtained by the arrangement of bonding electrons only.
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For the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5 →> NO₂+ NO3_(K1)
NO2 + NO3 → N2O5 (k-1)
NO2 + NO3 →
→
NO2 + O2 + NO (K2)
NO + N2O5-
NO2 + NO2 + NO2 (K3)
d[N₂O5] __2k‚k₂[N2O5]
Indicate whether the following rate expression is acceptable:
dt
k₁₁+ k₂
Consider the following decomposition reaction of N2O5(g):
For the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5 → NO2 + NO3 (K1)
NO2 + NO3 → N2O5 (k-1)
NO2 + NO3 → NO2 + O2 + NO (K2)
NO + N2O5 → NO2 + NO2 + NO2 (K3)
Indicate whether the following rate expression is acceptable:
d[N2O5]
= -k₁[N₂O₂] + K¸₁[NO₂][NO3] - K¸[NO₂]³
dt
In a reaction of A + B to give C, another compound other than A, B or C may appear in the kinetic equation.
Chapter 12 Solutions
EBK INTRODUCTORY CHEMISTRY
Ch. 12 - Prob. 1CECh. 12 - Prob. 2CECh. 12 - Prob. 3CECh. 12 - Prob. 4CECh. 12 - Prob. 5CECh. 12 - Prob. 6CECh. 12 - Prob. 7CECh. 12 - Prob. 8CECh. 12 - Prob. 9CECh. 12 - Prob. 10CE
Ch. 12 - Prob. 11CECh. 12 - Prob. 12CECh. 12 - Prob. 1KTCh. 12 - Prob. 2KTCh. 12 - Prob. 3KTCh. 12 - Prob. 4KTCh. 12 - Prob. 5KTCh. 12 - Prob. 6KTCh. 12 - Prob. 7KTCh. 12 - Prob. 8KTCh. 12 - Prob. 9KTCh. 12 - Prob. 10KTCh. 12 - Prob. 11KTCh. 12 - Prob. 12KTCh. 12 - Prob. 13KTCh. 12 - Prob. 14KTCh. 12 - Prob. 15KTCh. 12 - Prob. 16KTCh. 12 - Prob. 17KTCh. 12 - Prob. 18KTCh. 12 - Prob. 19KTCh. 12 - Prob. 20KTCh. 12 - Prob. 21KTCh. 12 - Prob. 22KTCh. 12 - Prob. 23KTCh. 12 - Prob. 24KTCh. 12 - Prob. 25KTCh. 12 - Prob. 26KTCh. 12 - Prob. 27KTCh. 12 - Prob. 28KTCh. 12 - Prob. 29KTCh. 12 - Prob. 1ECh. 12 - Prob. 2ECh. 12 - Prob. 3ECh. 12 - Prob. 4ECh. 12 - Prob. 5ECh. 12 - Prob. 6ECh. 12 - Prob. 7ECh. 12 - Prob. 8ECh. 12 - Prob. 9ECh. 12 - Prob. 10ECh. 12 - Prob. 11ECh. 12 - Prob. 12ECh. 12 - Prob. 13ECh. 12 - Prob. 14ECh. 12 - Prob. 15ECh. 12 - Prob. 16ECh. 12 - Prob. 17ECh. 12 - Prob. 18ECh. 12 - Prob. 19ECh. 12 - Prob. 20ECh. 12 - Prob. 21ECh. 12 - Prob. 22ECh. 12 - Prob. 23ECh. 12 - Prob. 24ECh. 12 - Prob. 25ECh. 12 - Prob. 26ECh. 12 - Prob. 27ECh. 12 - Prob. 28ECh. 12 - Prob. 29ECh. 12 - Prob. 30ECh. 12 - Prob. 31ECh. 12 - Prob. 32ECh. 12 - Prob. 33ECh. 12 - Prob. 34ECh. 12 - Prob. 35ECh. 12 - Prob. 36ECh. 12 - Prob. 37ECh. 12 - Prob. 38ECh. 12 - Prob. 39ECh. 12 - Prob. 40ECh. 12 - Prob. 41ECh. 12 - Prob. 42ECh. 12 - Prob. 43ECh. 12 - Prob. 44ECh. 12 - Prob. 45ECh. 12 - Prob. 46ECh. 12 - Prob. 47ECh. 12 - Prob. 48ECh. 12 - Prob. 49ECh. 12 - Prob. 50ECh. 12 - Prob. 51ECh. 12 - Prob. 52ECh. 12 - Prob. 53ECh. 12 - Prob. 54ECh. 12 - Prob. 55ECh. 12 - Prob. 56ECh. 12 - Prob. 57ECh. 12 - Prob. 58ECh. 12 - Prob. 59ECh. 12 - Prob. 60ECh. 12 - Prob. 61ECh. 12 - Prob. 62ECh. 12 - Prob. 63ECh. 12 - Prob. 64ECh. 12 - Prob. 65ECh. 12 - Prob. 66ECh. 12 - Prob. 67ECh. 12 - Prob. 68ECh. 12 - Prob. 69ECh. 12 - Prob. 70ECh. 12 - Prob. 71ECh. 12 - Prob. 72ECh. 12 - Prob. 73ECh. 12 - Prob. 74ECh. 12 - Prob. 75ECh. 12 - Prob. 76ECh. 12 - Prob. 77ECh. 12 - Prob. 78ECh. 12 - Prob. 79ECh. 12 - Prob. 80ECh. 12 - Prob. 81ECh. 12 - Prob. 82ECh. 12 - Prob. 83ECh. 12 - Prob. 84ECh. 12 - Prob. 85ECh. 12 - Prob. 86ECh. 12 - Prob. 87ECh. 12 - Prob. 88ECh. 12 - Prob. 89ECh. 12 - Prob. 90ECh. 12 - Prob. 91ECh. 12 - Prob. 92ECh. 12 - Prob. 93ECh. 12 - Prob. 94ECh. 12 - Prob. 95ECh. 12 - Prob. 96ECh. 12 - Prob. 97ECh. 12 - Prob. 98ECh. 12 - Prob. 99ECh. 12 - Prob. 100ECh. 12 - Prob. 1STCh. 12 - Prob. 2STCh. 12 - Prob. 3STCh. 12 - Prob. 4STCh. 12 - Prob. 5STCh. 12 - Prob. 6STCh. 12 - Prob. 7STCh. 12 - Prob. 8STCh. 12 - Prob. 9STCh. 12 - Prob. 10STCh. 12 - Prob. 11STCh. 12 - Prob. 12STCh. 12 - Prob. 13STCh. 12 - Prob. 14STCh. 12 - Prob. 15STCh. 12 - Prob. 16STCh. 12 - Prob. 17STCh. 12 - Prob. 18ST
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Similar questions
- For the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 →> NO₂+ NO3_(K1) NO2 + NO3 → N2O5 (k-1) NO2 + NO3 → → NO2 + O2 + NO (K2) NO + N2O5- NO2 + NO2 + NO2 (K3) d[N₂O5] __2k‚k₂[N2O5] Indicate whether the following rate expression is acceptable: dt k₁₁+ k₂arrow_forwardGiven the reaction R + Q → P, indicate the rate law with respect to R, with respect to P and with respect to P.arrow_forwardSteps and explanations. Also provide, if possible, ways to adress this kind of problems in general.arrow_forward
- k₁ Given the reaction A B, indicate k-1 d[A] (A). the rate law with respect to A: (B). the rate law with respect to B: d[B] dt dtarrow_forwardk₁ Given the reaction R₂ R + R, indicate k-1 (A). the rate law with respect to R2: (B). the rate law with respect to R: d[R₂] dt d[R] dtarrow_forwardGiven the reaction R+ Q → P, indicate (A). the rate law with respect to P: (B). the rate law with respect to R: (C). the rate law with respect to Q: d[P] dt d[R] dt d[Q] dtarrow_forward
- The reaction for obtaining NO2 from NO and O2 has the rate equation: v = k[NO]2[O2]. Indicate which of the following options is correct.(A). This rate equation is inconsistent with the reaction consisting of a single trimolecular step.(B). Since the overall order is 3, the reaction must necessarily have some trimolecular step in its mechanism.(C). A two-step mechanism: 1) NO + NO ⇄ N2O2 (fast); 2) N2O2 + O2 → NO2 + NO2 (slow).(D). The mechanism must necessarily consist of three unimolecular elementary steps with very similar rate constants.arrow_forwarda. What is the eluent used in the column chromatography here (a “silica plug filtration” is essentially a very short column)? b. The spectroscopy of compound 5b is described in the second half of this excerpt, including 1H-NMR and 13C-NMR (which you will learn about in CHEM 2412L), MS (which you will learn about later in CHEM 2411L) and IR. One of the IR signals is at 3530 cm-1. What functional group does this indicate might be present in compound 5b?arrow_forwardSteps and explanations. Also provide, if possible, ways to adress this kind of problems in general.arrow_forward
- a. The first three lines of this procedure describe the reaction used to make compound 5b. In the fourth line, hexane and sodium bicarbonate are added. What organic lab technique is being used here? b. What is the purpose of the Na2SO4? c. What equipment would you use to “concentrate [a solution] under reduced pressure”?arrow_forwardWhen N,N-dimethylaniline is treated with bromine both the ortho and para products are observed. However when treated with a mixture of nitric acid and sulfuric acid only the meta product is observed. Explain these results and support your answer with the appropriate drawings *Hint amines are bases* N HNO3 H2SO4 N NO2 N Br2 N Br + N 8-8-8 FeBr3 Brarrow_forwardDraw a mechanism that explains the formation of compound OMe SO3H 1. Fuming H2SO4arrow_forward
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