Question

For a general chemical equation

A+B⇌C+D

the equilibrium constant can be expressed as a ratio of the concentrations:

*K*c=[C][D]/[A][B]

If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows:

forward rate=*k*f[A][B]

reverse rate=*k*r[C][D]

where *k*f and *k*r are the forward and reverse
rate constants, respectively. When equilibrium is reached, the
forward and reverse rates are equal:

*k*f[A][B]=*k*r[C][D]

Thus, the rate constants are related to the equilibrium constant in the following manner:

*K*c=*k*f/*k*r=[C][D]/[A][B]

1.For a certain reaction, Kc = 1.43×1010 and kf= 6.45×10−4 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.

2.For a different reaction, *K**c* =
1.88×10^{7}, *k*f=8.92×105s−1, and *k*r=
4.75×10^{−2} s−1 . Adding a catalyst increases the forward
rate constant to 1.34×10^{7} s−1 . What is the new value of
the reverse reaction constant, *k*r, after adding
catalyst?

3.Yet another reaction has an equilibrium constant
*K*c=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? (increase/decrease/not change)

Answer #1

1.

K_{c} = 1.43 ×10^{10} and k_{f}=
6.45×10^{−4} M^{−2}s^{−1}

k_{r} = k_{f}/K_{c} = (6.45
×10^{−4} M^{−2}s^{−1}) /
(1.43×10^{10}) = 4.51 ×10^{−14}
M^{−2}s^{−1}

2.

K_{c} = 1.88×10^{7} and k_{f}= 1.34×10^{7} s^{−1}

k_{r} = k_{f}/K_{c} = (1.34
×10^{7} s^{−1}) / (1.88×10^{7}) = 7.12
×10^{−1} s^{−1}

3. Raising the
temperature to 200
^{∘}C favors the reverse reaction (endothermic).

Equilibrium constant changes if we change the temperature of the
system. K_{c} is constant at constant temperature, but they
vary as the temperature changes. On increasing temperature in is an
exothermic reaction, the value of K_{c} decreases.

To understand the relationship between the equilibrium constant
and rate constants.
For a general chemical equation
A+B⇌C+D
the equilibrium constant can be expressed as a ratio of the
concentrations:
Kc=[C][D][A][B]
If this is an elementary chemical reaction, then there is a
single forward rate and a single reverse rate for this reaction,
which can be written as follows:
forward ratereverse rate==kf[A][B]kr[C][D]
where kf and kr are the forward and reverse
rate constants, respectively. When equilibrium is reached, the
forward and...

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For a certain reaction, Kc = 8.85×1010 and kf= 7.52×10−2 M−2⋅s−1
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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.
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For a different reaction, Kc =
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a. For a certain reaction, Kc=
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Calculate the
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kf=4.58×103s?1, and kr= 0.684 s?1 . Adding a
catalyst increases the forward rate constant to 1.04×106
s?1 . What is the new value of the reverse reaction constant,
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Express your answer numerically in inverse seconds.
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For a different reaction, Kc = 1.92×103, kf=6.92×104s−1, and kr=
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kr, given that the reverse reaction is of the same molecularity as
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Express your answer with the appropriate units. Include explicit
multiplication within units, for example, to enter M−2⋅s−1 include
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The reversible chemical reaction
A+B⇌C+D
has the following equilibrium constant:
Kc=[C][D]/[A][B]=2.5
1. Initially, only A and B are present, each at 2.00 M.
What is the final concentration of A once equilibrium is
reached?
2.What is the final concentration of D at equilibrium if the
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Consider the gas-phase hydration of hexafluoroacetone,
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(CF3)2CO(g)+H2O(g)⇌krkf(CF3)2C(OH)2(g)
At 76 ∘C, the forward and reverse rate constants are
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What is the value of the equilibrium constant Kc?

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A+B⇌C+D
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B:What is the final concentration of D at equilibrium if the
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M ?
Please include all steps!

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