Step 7 of 8 Substitute the value of [Br] from Step 3 and the value of [H] from Step 5 into the expression for d[HBr] dt s and simplify. (Use the following as necessary: [Br2], [H2], [HBr], k1, k2, k3, ka, and ks.) from Step
Analyzing Infrared Spectra
The electromagnetic radiation or frequency is classified into radio-waves, micro-waves, infrared, visible, ultraviolet, X-rays and gamma rays. The infrared spectra emission refers to the portion between the visible and the microwave areas of electromagnetic spectrum. This spectral area is usually divided into three parts, near infrared (14,290 – 4000 cm-1), mid infrared (4000 – 400 cm-1), and far infrared (700 – 200 cm-1), respectively. The number set is the number of the wave (cm-1).
IR Spectrum Of Cyclohexanone
It is the analysis of the structure of cyclohexaone using IR data interpretation.
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Interpretation of anisole using IR spectrum obtained from IR analysis.
IR Spectroscopy
Infrared (IR) or vibrational spectroscopy is a method used for analyzing the particle's vibratory transformations. This is one of the very popular spectroscopic approaches employed by inorganic as well as organic laboratories because it is helpful in evaluating and distinguishing the frameworks of the molecules. The infra-red spectroscopy process or procedure is carried out using a tool called an infrared spectrometer to obtain an infrared spectral (or spectrophotometer).
![H2(g)+Br2(g)
- 2HB1(g)
- 2 HBr(g)
H2(g) + Br2(g)
H_2(g)+Br_2(g)-->2HBr(g)
Correct.
Step 2 of 8
Assuming that the reaction chain proceeds according to the elementary steps below, apply the steady state approximation to both the H and Br atoms. (Use the following as necessary: [Br], [Br2], [H], [H2], and [HBr], or
enter 0 if the term does not apply.)
elementary reaction
rate constant
reaction type
Br2 - 2 Br
k1
chain initiation
Br + H2 - HBr + H
K2
chain propagation
H + Brz - HBr + Br
k3
chain propagation
H + HBr - H2 + Br
k4
chain inhibition
Br + Br - Br2
k5
chain termination
d[H]
dt
= 0 = k2l LBr ]| H,
[Br][H2]
k3
[H] Br2
[H][Br2]
- kal [H]HBr
|[H][HBr]
d[Br]
= 0 = k1( 2| Br2
2[Brz]
dt
Br][ H2
- k2
[Br][H2]
+ k3
H Br.
[H][Brz]
+ ka [H][HBr]
[H][HBr]
- ksl
[Br]2
[Br]²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14e51fbd-8bc7-4604-ba1b-5c934c5c4cfb%2F3e56b7d6-e071-4722-b9a9-034af2c0945c%2Fn8mcqtq_processed.png&w=3840&q=75)
![Step 3 of 8
Sum the equations for
dt
d[H]
d[Br]
and solve for [Br]. (Use the following as necessary: [Br2], [H], [H2], [HBr], k1, k2, k3, k4, and k5.)
and
dt
2k,
[Br2]
2k; [Br]
[Br] =
k5
Step 4 of 8
d[H]
from Step 2, solve for [H]. (Use the following as necessary: [Br], [Br2], [H2], [HBr], k1, k2, k3, k4, and k5.)
dt
Using the equation for
ky[ Br ][H,]
k3 Br, + k4
[HBr]
[H] =
k2[Br][H2]
k3[Bra] + k4[HBr]
Step 5 of 8
Substitute the equation for [Br] from Step 3 into the equation for [H] from Step 4 and simplify. (Use the following as necessary: [Br2], [H2], k1, k2, k3, k4, and k5.)
2k, Br2
k2 V
[#=]
2k1
k2
V[Bra]|H2]
K5
[H]
k3[Br2] + k4[HBr]
Step 6 of 8
Use the given elementary steps from Step 2 to define HDIL. (Use the following as necessary: [Br], [Br2], [H2], [HBr], k1, k2, k3, k4, and k5.)
dt
d[HBr]
[Br) k2 H,
dt
ka[H2]
+ [HI k3 Br,
– k,[HBr
|(k3[Bra]) – (k4[HBr])
Step 7 of 8
Substitute the value of [Br] from Step 3 and the value of [H] from Step 5 into the expression for ADI from Step 6 and simplify. (Use the following as necessary: [Br2], [H2], [HBr], k1, k2, k3, k4, and k5.)
dt
d[HBr]
dt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14e51fbd-8bc7-4604-ba1b-5c934c5c4cfb%2F3e56b7d6-e071-4722-b9a9-034af2c0945c%2Fsc4zya_processed.png&w=3840&q=75)
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