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lec11_2011

# lec11_2011 - 22.4 Reactions approaching equilibrium Case 1...

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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 = 0, one has [A] + [B] = [A] 0 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] 0 – [A] eq = ] [ ' ] [ ] [ B k A k dt A d + - = = - + - = ]) [ ] ([ ' ] [ ] [ A A k A k dt A d 0 0 ] [ ' ] )[ ' ( A k A k k + + - 0 ] [ ' ' ) ' ( A k k ke k t k k + + + - ' ] [ ' k k A k + 0 ' ] [ k k A k + 0

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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 = ] [ τ / t e x x - = 0 b a k k + = τ 1 τ

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Example 22.4: The H 2 O( l ) ↔ H + (aq) + OH - (aq) equilibrium relaxes in 37 μs at 298 K and pKw = 14.0. Calculate the rate constants for the forward and backward reactions.
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lec11_2011 - 22.4 Reactions approaching equilibrium Case 1...

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