The first-order rate constant for the conversion of cyclobutane to ethylene at 1000.°C is 101 s²¹. C₂H82 C₂H4 A. What is the half-life of the reaction at 1000."C? B. If you started with 3.350 g of cyclobutane, how long would it take to consume 0.4020 g of it? C. How much of a 2.50 g sample of cyclobutane would remain after 0.03430 s?
The first-order rate constant for the conversion of cyclobutane to ethylene at 1000.°C is 101 s²¹. C₂H82 C₂H4 A. What is the half-life of the reaction at 1000."C? B. If you started with 3.350 g of cyclobutane, how long would it take to consume 0.4020 g of it? C. How much of a 2.50 g sample of cyclobutane would remain after 0.03430 s?
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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DETAILS
The first-order rate constant for the conversion of cyclobutane to ethylene at 1000.°C is 101 s ¹.
C4H8 → 2 C₂H4
A. What is the half-life of the reaction at 1000."C? [
B. If you started with 3.350 g of cyclobutane, how long would it take to consume 0.4020 g of it?
C. How much of a 2.50 g sample of cyclobutane would remain after 0.03430 s?
Useful Information:
1.013 bar = 760 torr = 1 atm = 760 mm Hg
h = 6.62607x10-34 3.s
c = 2.99792 x 108 m/s
R = 8.314472 3/mol.K
NA = 6.02214 x 1023 1/mol
R = 0.08205746 L.atm/mol.K
F = 9.648534 x 104 c/mol
Equations:
K= K (RT)An
AG=AHⓇ - TASⓇ
MY NOTES
pH = pk +log.
E = E
[A]
[HA]
7=45
0.05916
n
ASK YOUR TEACHER PRACTICE ANOTHER
log Qrxn
AG=-RTINK
AG=-nFE
1
[A]t
In(2)
k
[A] = [A]okt
1
=kt +
[A]
9
[A]
t=
k[A]
[A]
2k
-kt
-Ea
k Ae RT](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc18503d3-c27d-4700-ac99-ac57f99c44ef%2Fdfa01aba-e96e-4fec-b2e9-19ba091c73d3%2Forhrn2j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:7.
DETAILS
The first-order rate constant for the conversion of cyclobutane to ethylene at 1000.°C is 101 s ¹.
C4H8 → 2 C₂H4
A. What is the half-life of the reaction at 1000."C? [
B. If you started with 3.350 g of cyclobutane, how long would it take to consume 0.4020 g of it?
C. How much of a 2.50 g sample of cyclobutane would remain after 0.03430 s?
Useful Information:
1.013 bar = 760 torr = 1 atm = 760 mm Hg
h = 6.62607x10-34 3.s
c = 2.99792 x 108 m/s
R = 8.314472 3/mol.K
NA = 6.02214 x 1023 1/mol
R = 0.08205746 L.atm/mol.K
F = 9.648534 x 104 c/mol
Equations:
K= K (RT)An
AG=AHⓇ - TASⓇ
MY NOTES
pH = pk +log.
E = E
[A]
[HA]
7=45
0.05916
n
ASK YOUR TEACHER PRACTICE ANOTHER
log Qrxn
AG=-RTINK
AG=-nFE
1
[A]t
In(2)
k
[A] = [A]okt
1
=kt +
[A]
9
[A]
t=
k[A]
[A]
2k
-kt
-Ea
k Ae RT
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