Question

**Part A**

The activation energy of a certain reaction is 48.4 kJ/mol . At 24 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast

T2 = |
35 ∘C |

**Part B**

Given that the initial rate constant is 0.0190s−1 at an initial temperature of 24 ∘C , what would the rate constant be at a temperature of 190. ∘C for the same reaction described in Part A?

*k*2 =?

Answer #1

Apply Ahrrenius

ln(k2/k1) = E/R*(1/T1-1/T2)

ln(2k1/k1) = (48400)/(8.314)*(1/(24+273) - 1/T2)

Solve for T2

1/T = (ln(2k1/k1)*8.314/(48400) - 1/(24+273)) = -0.00324

T = -(-0.00324^-1) = 308.6419 K

T = 308.6419-373 = 35.6419 °C

B)

E = 48400 and R = 8.314

if rate K = 0.019 and T = 24 °C = 299 K

then

find k2 at T = 190 ° C = 463

ln(k2/k1) = E/R*(1/T1-1/T2)

ln(k2/(0.019)) = 48400/8.314*(1/(299)-1/463)

solve for k2

ln(k2/(0.019)) = 6.896472

k2 = 0.019*exp(6.896472) = 18.7868 s^-1

Part A
The activation energy of a certain reaction is 30.5 kJ/mol . At
30 ∘C , the rate constant is 0.0180s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast?
Express your answer with the appropriate units.
T2 =
Part B
Given that the initial rate constant is 0.0180s−1 at an initial
temperature of 30 ∘C , what would the rate constant be
at a temperature of 170. ∘C for the same reaction
described in Part A?
Express...

Part A: The activation energy of a certain
reaction is 43.5 kJ/mol . At 23 ∘C , the rate constant is
0.0180s−1. At what temperature in degrees Celsius would this
reaction go twice as fast?
Part B: Given that the initial rate constant is
0.0180s−1 at an initial temperature of 23 ∘C , what would the rate
constant be at a temperature of 190. ∘C for the same reaction
described in Part A?

Part A: The activation energy of a certain reaction is 42.3
kJ/mol . At 29 ∘C , the rate constant is 0.0170s−1 . At what
temperature in degrees Celsius would this reaction go twice as
fast?
Part B: Given that the initial rate constant is 0.0170s−1 at an
initial temperature of 29 ∘C , what would the rate constant be at a
temperature of 120. ∘C for the same reaction described in Part
A?

Part A:
The activation energy of a certain reaction is 42.8 kJ/mol . At
28 ∘C , the rate constant is 0.0190s−1. At what temperature in
degrees Celsius would this reaction go twice as fast?
Part B:
Given that the initial rate constant is 0.0190s−1 at an initial
temperature of 28 ∘C , what would the rate constant be
at a temperature of 150 ∘C for the same reaction
described in Part A?

Part A The activation energy of a certain reaction is 30.6
kJ/mol . At 22 ∘C , the rate constant is 0.0110s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast? Express your answer with the appropriate units. Hints T2 = 39
∘C SubmitMy AnswersGive Up Correct Part B Given that the initial
rate constant is 0.0110s−1 at an initial temperature of 22 ∘C ,
what would the rate constant be at a temperature of 170....

The activation energy of a certain reaction is 35.1 kJ/mol . At
25 ∘C , the rate constant is 0.0160s−1. At what temperature in
degrees Celsius would this reaction go twice as fast?
Given that the initial rate constant is 0.0160s−1 at an initial
temperature of 25 ∘C , what would the rate constant be
at a temperature of 200. ∘C for the same reaction
described in Part A?

Part A:
The activation energy of a certain reaction is 50.0 kJ/mol . At
25 ∘C , the rate constant is 0.0110s−1. At
what temperature in degrees Celsius would this reaction go twice as
fast?
Express your answer with the appropriate units.
Part B:
Given that the initial rate constant is 0.0110s−1 at
an initial temperature of 25 ∘C , what would
the rate constant be at a temperature of
140. ∘C for the same reaction described in
Part A?
Express your...

Part A
The activation energy of a certain reaction is 36.4 kJ/mol . At
21 ∘C , the rate constant is 0.0160s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast?
Express your answer with the appropriate units.
Hints
Part B
Given that the initial rate constant is 0.0160s−1 at an initial
temperature of 21 ∘C , what would the rate constant be
at a temperature of 130. ∘C for the same reaction
described in Part A?
Express your...

Part A
The activation energy of a certain reaction is 43.9 kJ/mol . At
28 ∘C , the rate constant is 0.0130s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast?
Express your answer with the appropriate units.
Part B
Given that the initial rate constant is 0.0130s−1 at an initial
temperature of 28 ∘C , what would the rate constant be
at a temperature of 100. ∘C for the same reaction
described in Part A?
Express your answer...

A. The activation energy of a certain reaction is 42.0 kJ/mol .
At 29 ∘C , the rate constant is 0.0190s-1. At what
temperature in degrees Celsius would this reaction go twice as
fast? Express your answer with the appropriate units.
B. Given that the initial rate
constant is 0.0190s-1 at an initial temperature of 29 C
, what would the rate constant be at a temperature of
120. ∘C for the same reaction described in Part
A?
Express your answer...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 3 minutes ago

asked 5 minutes ago

asked 6 minutes ago

asked 7 minutes ago

asked 12 minutes ago

asked 26 minutes ago

asked 28 minutes ago

asked 30 minutes ago

asked 33 minutes ago

asked 38 minutes ago

asked 39 minutes ago

asked 42 minutes ago