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posted by  ss83 on 7/7/2008 9:28:17 PM  |  status: Live  

Chemical Equilibrium and Chemical Kinetics

Course Textbook Chapter Problem
General Chemistry N/A 14 N/A
Question Details:

Learning Goal: To understand the relationship between the equilibrium constant and rate constants.

For a general chemical equation

\rm A+B \rightleftharpoons C+D

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

K_{\rm c}={\rm \frac{[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:

\matrix {\hfill \mbox{forward rate}&=&k_{\rm f}{\rm [A][B]}\hfill \cr \hfill \mbox{reverse rate}&=&k_{\rm r}{\rm [C][D]}\hfill \cr}

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_{\rm f} {\rm[A][B]}=k_{\rm r} {\rm [C][D]}

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

K_{\rm c}=\frac{k_{\rm f}}{k_{\rm r}}={\rm \frac{[C][D]}{[A][B]}}



For a certain reaction, K_c=154 and k_{\rm f}=97.6 <units>{\it M}^{-2}\cdot s^{-1}</units>. Calculate the value of the reverse rate constant, k_r.
Express your answer numerically in inverse seconds.
  k_r  =
  M^{-2}\cdot \rm s^{-1}
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posted by AJ <3 on 7/7/2008 10:23:39 PM  |  status: Live
Asker's Rating: Lifesaver   
ss83's comment:
"you helped me out a lot with understanding this problem, thank you!"
Response Details:
Seems like I answered a question similar to this.

By this relationship:

Kc = kf / kr

So from the given information:

K_c=154
k_{\rm f}=97.6 <units />{\it M}^{-2}\cdot s^{-1}</units>
kr = ?                            <--- This is the unknown rate constant we are trying to find.

So using the equation from the beginning, Kc = kf / kr, plug everything in and solve for kr:

Kc = kf / kr
(kr) (Kc) = kf
kr = kf / Kc
    = 97.6 M^{-2}\cdot \rm s^{-1}  / 154
    = .634 M^{-2}\cdot \rm s^{-1} 

Please ask for further assistance.
Please ask for further assistance. AJ
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