Question

A reaction has a rate constant of 1.25×10−4 s−1 at 29 ∘C and 0.228 s−1 at 79 ∘C . Part A Determine the activation barrier for the reaction. Express your answer in units of kilojoules per mole. Part B What is the value of the rate constant at 18 ∘C ? Express your answer in units of inverse seconds.

Answer #1

given T1= 29 deg.c= 29+273= 302K, T2= 79deg.c= 79+273= 352K,

K1= rate constant at T1= 1.25*10^{-4}/s and K2= rate
constant at T2= 0.228/s

From Arhenius equation, ln (K2/K1)= (Ea/R)*(1/T1-1/T2)

Ea= activation energy , R= 8.314 J/mole.K

ln (0.228/1.25*10^{-4})= (Ea/8.314)*(1/302-1/352)

6.82= (Ea/R)*0.000437, Ea= 6.82*8.314/0.000437 J/ mole =120552 J/mole= 120.55 Kj/mole

at T3=18 deg.c= 18+273= 291K, Writing Arhenius equations for T3 and T1 and noting that K3= rate constant at T3.

ln (K3/K1)= (Ea/R)*(1/302-1/291)=(120.55*1000/8.314)*(-0.00013)

K3/K1=0.163, K3=
0.163*1.25*10^{-4}/s=2.04*10^{-5}/s

A reaction has a rate constant of 1.26×10−4 s−1 at 28 ∘C and
0.233 s−1 at 79 ∘C .
1.Determine the activation barrier for the reaction. Express
your answer in units of kilojoules per mole.
2. What is the value of the rate constant at 17 ∘C ? Express
your answer in units of inverse seconds.

A reaction has a rate constant of 1.21×10−4 s−1 at
26 ∘C and 0.229 s−1 at 75 ∘C .
You may want to reference (Page) section 13.5 while completing
this problem
Part A
Determine the activation barrier for the reaction.
(Express your answer in units of kilojoules per mole.)
Part B
What is the value of the rate constant at 18 ∘C ?
(Express your answer in units of inverse seconds.)

A reaction has a rate constant of 1.21×10^−4 s^(−1) at 26 ∘C and
0.229 s^(−1) at 79 ∘C .
A) Determine the activation barrier for the reaction.
B) What is the value of the rate constant at 16 ∘C ?

A reaction has a rate constant of 1.25×10−2 /s at
400. Kand 0.686 /s at 450. K.
1. Determine the activation barrier for the reaction.
2.What is the value of the rate constant at 425 K?

A reaction has a rate constant of 1.22×10−4 s−1 at
26 ∘C and 0.234 s−1 at 75 ∘C .
A) Determine the activation barrier for the reaction.
B)What is the value of the rate constant at 15 ∘C
?

A reaction has a rate constant of 1.28×10−4 s−1 at
25 ∘C and 0.226 s−1 at 78 ∘C .
A) Determine the activation barrier for the reaction.
B) What is the value of the rate constant at 16 ∘C
?

The rate constant of a chemical reaction increased from 0.100
s−1 to 3.10 s−1 upon raising the temperature from 25.0 ∘C to 51.0
∘C .
Part A
Calculate the value of (1/T2−1/T1) where
T1 is the initial temperature and T2 is the final
temperature.
Express your answer numerically.
Part B
Calculate the value of ln(k1/k2) where
k1 and k2 correspond to the rate constants at the
initial and the final temperatures as defined in part A.
Express your answer numerically....

The rate constant of a chemical reaction increased from 0.100
s−1 to 2.60 s−1 upon raising the temperature from 25.0 ∘C to 45.0
∘C .
I solved
(1T2−1T1) =
−2.11×10−4
K−1
In (k1/k2) = -3.26
But, I'm having problems on this question:
What is the activation energy of the
reaction?
Express your answer numerically in kilojoules per
mole.
It would be great if you could show all of your work. I've been
trying to figure out this problem for a...

Part A
A certain first-order reaction (A→products) has a rate constant
of 7.20×10−3 s−1 at 45 ∘C. How many minutes does it take
for the concentration of the reactant, [A], to drop to 6.25% of the
original concentration?
Express your answer with the appropriate units.
Answer:
6.42 min
Part B
A certain second-order reaction (B→products) has a rate constant
of 1.35×10−3M−1⋅s−1 at 27 ∘Cand an initial
half-life of 236 s . What is the concentration of the reactant B
after...

Part A
The rate constant for a certain reaction is k = 3.70×10−3 s−1 .
If the initial reactant concentration was 0.200 M, what will the
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Part B
A zero-order reaction has a constant rate of
2.00×10−4M/s. If after 65.0 seconds the
concentration has dropped to 3.50×10−2M, what
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Express your answer with the appropriate units.

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