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Unformatted text preview: 22.4 Reactions approaching equilibrium Case 1: First order reactions : A B v = k [A] B A v = k [B] the net rate change for A is therefore if [B] = 0, one has [A] + [B] = [A] at all time. the integrated solution for the above equation is [A] = As t , the concentrations reach their equilibrium values: [A] eq = [B] eq = [A] [A] eq = ] [ ' ] [ ] [ B k A k dt A d + = = + = ]) [ ] ([ ' ] [ ] [ A A k A k dt A d ] [ ' ] )[ ' ( A k A k k + + ] [ ' ' ) ' ( A k k ke k t k k + + + ' ] [ ' k k A k + ' ] [ k k A k + The equilibrium constant can be calculated as K = thus: In a simple way, at the equilibrium point there will be no net change and thus the forward reaction will be equal to the reverse reaction: k [A] eq = k [B] eq thus the above equation bridges the thermodynamic quantities and reaction rates through equilibrium constant . For a general reaction scheme with multiple reversible steps: eq eq A B ] [ ] [ ' k k K = ' ] [ ] [ k k A B eq eq = ... ' ' = 2 2 1 1 k k k k K Determining rate constants with relaxation method After applying a perturbation, the system ( A B) may have a new equilibrium state. Assuming the distance between the current state and the new equilibrium state is x, one gets [A] = [A] eq + x; [B] = [B] eq x; Because one gets dx/dt =  ( k a + k b ) x therefore is called the relaxation time x k k x B k A x k dt A d b a eq b eq a ) ( ) ] ([ ) ] [ ( ] [ + = + + = dt dx dt A d = ] [ / t e x x = b a k k + = 1 Example 22.4: Example 22....
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This note was uploaded on 02/14/2011 for the course SCIENCE 321 taught by Professor Aaa during the Spring '11 term at Windsor.
 Spring '11
 aaa
 Equilibrium

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