Chemical Equilibrium and Chemical Kinetics Part A For a certain reaction, Kc = 8

Question

Chemical Equilibrium and Chemical Kinetics

Part A

For a certain reaction, Kc = 8.85×1010 and kf= 7.52×102 M2s1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction. Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M2s1 include (multiplication dot) between each measurement.

Part B

For a different reaction, Kc = 1.70×1010, kf=6.63×105s1, and kr= 3.91×105 s1 . Adding a catalyst increases the forward rate constant to 2.06×108 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?

Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M2s1 include (multiplication dot) between each measurement.

Part C

Yet another reaction has an equilibrium constant Kc=4.32×105 at 25 C. It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to200 C , what will happen to the equilibrium constant?

The equilibrium constant will

Yet another reaction has an equilibrium constant  at 25 . It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200  , what will happen to the equilibrium constant?

Explanation / Answer

A) we knwo that

Kc = Kf / Kr

so

8.85 x 10^10 = 7.52 x 10-2 / Kr

Kr = 8.5 x 10-13

so

kr value is 8.5 x 10-13 M-2 s-1

B)

while adding catalyst KC value remains same

so

1.70 x 10^10 = 2.06 x 10^8 / Kr

Kr = 0.0153

so

Kr value is 0.0153 M-2 s-1

C)

we know that

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

now

T2 > T1

1 /T1 > 1 / T2

1 / T1 - 1/T2 > 0

also

for exothermic reaction Ea < 0

so

ln ( k2 / k1) < 0

so

k2 < k1

so

the equilibrium constant decreases

answer is

B) decrease

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.