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Equilibrium [3]

# Equilibrium [3] - Equilibrium Calculation Small K 2 N 2 g...

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1 6BH-Equilibrium (Basics) Keq Basics- 30 Equilibrium Calculation: Small K Since K is small, we should be able to ignore x when added to or subtracted from a value, as long as x is less than 5% of the value it is being added to or subtracted from –37 22 2 2 N ( ) O ( ) 2 N O( ) 2.0 10 c gg g K × ⎯⎯→ += ←⎯⎯ [ ] [] [ ] 2 2– 3 7 2 –37 NO 2.0 10 2.0 10 0.0482–2 0.0933– 0.0482 0.0933 c K xx × × == ≈= 6BH-Equilibrium (Basics) Keq Basics- 31 Equilibrium Calculation: Small K (cont.) Solution of this gives x = 3.3 × 10 –21 M Was neglecting x reasonable? Check to see if 2x < 5% of 0.0482 (since N 2 was limiting here): 0.05 × 0.0482 = 0.00241 0.00241 > 2(3.3 × 10 –21 ) = 6.6 × 10 –21 = 2 x Therefore, neglecting x was a reasonable approximation! So, at equilibrium: [N 2 ] 0.0482 M , [O 2 ] 0.0933 M , and [N 2 O] 6.6 × 10 –21 M

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2 6BH-Equilibrium (Basics) Keq Basics- 32 Intermediate K • An important case, primarily in biological systems where reactant and product concentrations can be within 1-2 orders of magnitude of K; calculations are done the same way for all cases; • Use an equilibrium table! • Binomials or polynomials WILL be involved! • Solution will employ the quadratic equation or successive approximative methods. 6BH-Equilibrium (Basics) Keq Basics- 33 Intermediate K : Quadratic Solution Calculate the [H 3 O + ] at equilibrium for a 0.025 M HSO 4 solution. K a = 1.20 × 10 –2 –– 2 + 42 4 3 HSO ( ) H O( ) SO ( H O ( ) . 0.025 0 0 –— + + . (0.025– ) aq l aq aq Init M xx x Eq xM ++ ⎯⎯→ ←⎯⎯ [] () +– 2 34 –2 4 2– 4 2 4 HO SO 1.20 10 HSO 0.025– 3.0 10 – 1.20 10 1.20 10 –3.0 10 0 c K x × ×× == = = += ⎡⎤ ⎣⎦ UGH!
3 6BH-Equilibrium (Basics) Keq Basics- 34 Intermediate K : Quadratic Solution Solve using the quadratic equation: 2 2 –– 4 ;based on 0 2a bba c xa x b x c ± =+ + = () ( ) ( )

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Equilibrium [3] - Equilibrium Calculation Small K 2 N 2 g...

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