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Reaction Quotient-Q- Or Trial K. The Keq is a constant- a number that does not change Increasing the Temperature of an endothermic equilibrium shifts.

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Presentation on theme: "Reaction Quotient-Q- Or Trial K. The Keq is a constant- a number that does not change Increasing the Temperature of an endothermic equilibrium shifts."— Presentation transcript:

1 Reaction Quotient-Q- Or Trial K

2 The Keq is a constant- a number that does not change Increasing the Temperature of an endothermic equilibrium shifts right and increases the Keq Increasing a [Reactant] shifts right but maintains the Keq Only temperature changes the Keq Changing the volume, pressure, or any concentration, does not change the Keq.

3 K trial How can you tell if a system is in equilibrium or not? Calculate a trial Keq or Q- reaction quotient. Use initial concentrations in the equilibrium expression to evaluate.

4 K eq KtKt How can you tell if a system is in equilibrium or not? Calculate a trial K eq. Put initial concentrations into the equilibrium expression and evaluate. If K t = K eq equilibrium If K t > K eq KtKt KtKt If K t < K eq Shifts right( to products) Shifts left (to reactants) 52035

5 Shifts right! 2NH 3(g) ⇄ N 2(g) + 3H 2(g) Keq = moles of NH 3, 15.0 moles of N 2, and 10.0 moles of H 2 are put in a 5.0 L container. Is the system in equilibrium and how will it shift if it is not? 2.0 M3.0 M2.0 M Kt= [N 2 ][H 2 ] 3 [NH 3 ] 2 =(3)(2) 3 = 6 (2) 2 Not in equilibrium Kt < Keq

6 Shifts left! 2NH 3(g) ⇄ N 2(g) + 3H 2(g) Keq = x moles of NH 3, 5.62 x moles of N 2, and 2.66 x moles of H 2 are put in a mL container. Is the system in equilibrium and how will it shift if it is not? 9.12 x M1.124 x M5.32 x M Kt= [N 2 ][H 2 ] 3 [NH 3 ] 2 =(1.124 x )(5.32 x ) 3 = 20.3 (9.12 x ) 2 Not in equilibrium Kt > Keq

7 3.If 4.00 moles of CO, 4.00 moles H 2 O, 6.00 moles CO 2, and 6.00 moles H 2 are placed in a 2.00 L container at 670 o C. CO (g) + H 2 O (g) ⇄ CO 2(g) + H 2(g) Keq = 1.0 Is the system at equilibrium? Calculate all equilibrium concentrations. Shifts left! Not in equilibrium Kt = (3)(3) (2)(2) = x +x -x -x 2.00 M 2.00 M 3.00 M 3.00 M x x x x K eq = [CO 2 ][H 2 ] [CO][H 2 O]

8 (3 - x) 2 (2 + x) 2 = x 2 + x 3 - x = 2 + x 1 =2x x = 0.50 M [CO 2 ]= [H 2 ] = = 2.50 M [CO]= [H 2 O] = = 2.50 M

9 Example F A 1.0 L reaction vessel contained 1.0 mol of SO 2, 4.0 mol of NO 2, 4.0 mol of SO 3 and 4.0 mol of NO at equilibrium according to SO 2(g) + NO 2(g) SO 3(g) + NO (g). If 3.0 mol of SO 2 is added to the mixture, what will be the new concentration of NO when equilibrium is re-attained?

10 SO 2(g) + NO 2(g) SO 3(g) + NO (g).

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