Question

For the bimolecular collision 2HI à H_{2} +
I_{2} the activation energy is 183 kJ/mol, d= 3.5 A and
the

steric factor is 0.44.

Predict the value of the rate constant as a function of temperature.

Answer #1

**Dear friend your answer is**

**Thank you**

The activation energy for the reaction H2(g) + I2(g) -->
2HI(g) is changed from 184 kJ/mol to 59.0 kJ/mol at 600 K by the
introduction of a Pt catalyst. Calculate the value of the ratio
rate(catalyzed)/rate(uncatalyzed).
A) 1.00
B) 7.62 x 10^10
C) 1.38
D) 0.321
E) none of these
I know the answer is B, But I want a detail solution.

The energy of activation for the reaction 2 HI → H2 + I2 is 180.
kJ·mol−1 at 556 K. Calculate the rate constant using the equation k
= Ae^(−Ea/RT). The collision diameter for HI is
3.5 ✕ 10−8 cm. Assume that the pressure is 1.00 atm.

The decomposition of hydrogen iodide, 2HI(g)= H2(g)+I2(g), has a
rate constant of 9.51*10-9 L/mol*s at 500K and Ea of 176 kJ/mol. At
what temperature will the rate constant be 1.10*10-5 L/mol*s?

a.)A certain reaction has an activation energy of 25.10 kJ/mol.
At what Kelvin temperature will the reaction proceed 7.00 times
faster than it did at 289 K?
b.A certain reaction has an enthalpy of ΔH = 39 kJ and an
activation energy of Ea = 51 kJ. What is the activation energy of
the reverse reaction?
c.)At a given temperature, the elementary reaction A<=> B in
the forward direction is the first order in A with a rate constant
of...

The activation energy, Ea for a particular reaction is
13.6 kj/mol. If the rate constant at 754 degrees celsius is
24.5/min at egat temperature in celsius will the rate constant be
12.7/min? r= 8.314j/mol • K

For the reaction H(g) + H2(g) ↔ (H-H-H)#→
H2(g) + H(g), activation energy Ea= 23 kJ/mol and the
preexponential factor A = 1.5 x 1010 L
mol-1s-1 at 298 K. Determine ∆H#,
∆S#, ∆G# and Kc#

The activation energy for a reaction is changed from 184 kJ/mol
to 59.5 kJ/mol at 600. K by the introduction of a catalyst. If the
uncatalyzed reaction takes about 2627 years to occur, about how
long will the catalyzed reaction take? Assume the frequency factor
A is constant and assume the initial concentrations are the
same.

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?

1) A first order reaction has an activation energy of 66.6
kJ/mol and a frequency factor (Arrhenius constant) of 8.78 x
1010 sec -1. Calculate the rate constant at
19 oC. Use 4 decimal places for your answer.
2) A first order reaction has a rate constant of 0.988 at 25
oC and 9.6 at 33 oC. Calculate the value of
the activation energy in KILOJOULES (enter answer to one decimal
place)

A)The activation energy of a certain reaction is 33.8 kJ/mol .
At 30 ∘C , the rate constant is 0.0170s−1. At what
temperature in degrees Celsius would this reaction go twice as
fast?
B)Given that the initial rate constant is 0.0170s−1 at an
initial temperature of 30 ∘C , what would the rate constant be at a
temperature of 200. ∘C for the same reaction described in Part
A?

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 28 minutes ago

asked 30 minutes ago

asked 33 minutes ago

asked 37 minutes ago

asked 49 minutes ago

asked 58 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago