Question

# Chemical Equilibrium and Chemical Kinetics Part A For a certain reaction, Kc = 8.85×1010 and kf=...

Chemical Equilibrium and Chemical Kinetics

Part A

For a certain reaction, Kc = 8.85×1010 and kf= 7.52×10−2 M−2⋅s−1 . 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 M−2⋅s−1 include ⋅ (multiplication dot) between each measurement.

Part B

For a different reaction, Kc = 1.70×1010, kf=6.63×105s−1, and kr= 3.91×10−5 s−1 . Adding a catalyst increases the forward rate constant to 2.06×108 s−1 . 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 M−2⋅s−1 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?

 A. increase. B. decrease. C. not change.

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

B) decrease

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