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posted by ss83 on 7/10/2008 8:14:37 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 different reaction, $K_{\rm c}=3.42×10^{3}$ , $k_{\rm f}=5.13×10^{3} <units>s^{-1}</units>$ , and $k_{\rm r}=1.50 <units>s^{-1}</units>$ . Adding a catalyst increases the forward rate constant to 8.31×10⁵ $s^{-1}$ . What is the new value of the reverse reaction constant, k_r