A reaction rate increases by a factor of 4000. in the presence of a catalyst at

Question

A reaction rate increases by a factor of 4000. in the presence of a catalyst at 45 degree C. The activation energy of the original uncatalyzed pathway is 125 kJ/mol. What is the activation energy of the new pathway (catalyzed), all other factors remaining constant. In (k_cat)/k_uncat = -1/RT [Ea_cat - Ea_uncat] For the reaction 2A + B + 2C rightarrow D + E, the following initial rate data was collected at constant temperature. Determine the correct rate law for this reaction. All units are arbitrary. From the attached graph, determine the freezing point of the solvent: FREEZING POINT ____________

Explanation / Answer

b)

T = 45 oC = (45 + 273) K = 318 K

Eauncat = 125 KJ/mol = 125000 J/mol

use:

ln (Kcat / Kuncat) = (-1/RT)*[EaCat - Eauncat]

ln (4000) = (-1/(8.314*318))*[EaCat - 125000]

[EaCat - 125000] = -21928

Eacat = 103072 J/mol

Eacat = 103.072 KJ/mol

Answer: 103.072 KJ/mol

7)

look at trial 1 and 2

all concentration except [A] is constant

[A] becomes 0.320/0.225 = 1.422

rate becomes = 0.0439/0.0217 = 2 times

since 1.422^2 = 2. So, order of A is 2

——————————————————

look at trial 1 and 3

all concentration except [B] is constant

[B] becomes 0.250/0.150 = 1.67

rate becomes = 0.0362/0.0217 = 1.67 times

So, order of B is 1

——————————————————

look at trial 1 and 4

all concentration except [C] is constant

[C] becomes 0.600/0.350 = 1.714

rate becomes = 0.01270/0.0217 = 0.585 times

let order of C be x

so,

0.585 = [1.714]^x

take ln on both sides

ln 0.585 = x* ln[1.714]

x = ln (0.585) / ln(1.714)

x = -1

So, order of C is 1

rate law is:

rate = k [A]^2 [B] [C]^-1

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