Chapter 15 Chemical Equilibrium

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Presentation transcript:

Chapter 15 Chemical Equilibrium Chapter 15 Chemical Equilibrium

What is Kp in terms of Kc for the following reaction ? What is Kp in terms of Kc for the following reaction ? 2NO (g) + O2 (g) 2NO2 (g) Kp = KcRT Kp = Kc/RT Kp = KcR/T Kp = Kc Kp = Kc/(RT)2

What is Kp in terms of Kc for the following reaction ? What is Kp in terms of Kc for the following reaction ? 2NO (g) + O2 (g) 2NO2 (g) Kp = KcRT Kp = Kc/RT Kp = KcR/T Kp = Kc Kp = Kc/(RT)2

What is the correct equilibrium constant expression for the reaction: P4 (s) + 6Cl2 (g) 4PCl3 (l)

What is the correct equilibrium constant expression for the reaction: P4 (s) + 6Cl2 (g) 4PCl3 (l)

Which accurately reflects the changes in concentration that will occur if O2 is added to disturb the equilibrium? 2NO (g) + O2 (g) 2NO2 (g) [NO] [O2] 1. Increase 2. Decrease 3. 4. 5. NO NO2

Which accurately reflects the changes in concentration that will occur if O2 is added to disturb the equilibrium? 2NO (g) + O2 (g) 2NO2 (g) [NO] [O2] 1. Increase 2. Decrease 3. 4. 5. NO NO2

PCl3 (g) + Cl2 (g) PCl5 (g) DH° = -87.9 kJ/mol Which of the following will result in an equilibrium shift to the right? PCl3 (g) + Cl2 (g) PCl5 (g) DH° = -87.9 kJ/mol Increase temperature/increase volume Increase temperature/decrease volume Decrease temperature/increase volume Decrease temperature/decrease volume None of the above PCl5

PCl3 (g) + Cl2 (g) PCl5 (g) DH° = -87.9 kJ/mol Which of the following will result in an equilibrium shift to the right? PCl3 (g) + Cl2 (g) PCl5 (g) DH° = -87.9 kJ/mol Increase temperature/increase volume Increase temperature/decrease volume Decrease temperature/increase volume Decrease temperature/decrease volume None of the above PCl5

What is the value of Kc for the reaction? What is the value of Kc for the reaction? 2CO(g) + O2(g) 2CO2(g) Kc = 5.0 x 1018 at 25 °C CO2(g) CO(g) + 1/2 O2(g) Kc = ?? at 25 °C Kc = 1/(5.0 x 1018)1/2 Kc = 1/(2.5 x 1018) Kc = -(5.0 x 1018)/2 Kc = -(5.0 x 1018)1/2 Kc = 2/(5.0 x 1018)1/2

What is the value of Kc for the reaction? What is the value of Kc for the reaction? 2CO(g) + O2(g) 2CO2(g) Kc = 5.0 x 1018 at 25 °C CO2(g) CO(g) + 1/2 O2(g) Kc = ?? at 25 °C Kc = 1/(5.0 x 1018)1/2 Kc = 1/(2.5 x 1018) Kc = -(5.0 x 1018)/2 Kc = -(5.0 x 1018)1/2 Kc = 2/(5.0 x 1018)1/2

For the gas-phase reaction A B the forward reaction rate is 3.0  104 s1 and the reverse reaction rate is 1.5  102 s1. What is the value of the equilibrium constant, Keq? 0.02 50 0.0004 2500

Correct Answer: 0.02 50 0.0004 2500

Does the following reaction represent a homogenous or a heterogenous equilbrium? C(s) + CO2(g) 2 CO(g) Homogenous Heterogeneous

Correct Answer: Homogenous Heterogeneous This is definitely a heterogeneous equilibrium because the reactants and products exist in two different phases, the solid and gaseous states.

2 Ag(s) + Zn2+(aq) 2 Ag+(aq) + Zn(s) Which of the following is the correct Keq for the equilibrium equation below? 2 Ag(s) + Zn2+(aq) 2 Ag+(aq) + Zn(s)

Correct Answer: [2Ag+][Zn] Keq = ___________ [2Ag][Zn2+] [Zn2+]

Correct Answer (cont.): Remember the equilibrium constant expression is given in terms of concentrations of products (raised to the exponent of their coefficients) divided by concentrations of reactants (raised to the exponent of their coefficients). Pure solids, liquids, and solvents are not included in the expression.

[ ] K B A D C = b a d c eq For the following hypothetical reaction: 2 Y(aq) + 3 Z(aq) A(aq) Calculate Keq given the following equilibrium concentrations: [A] = 0.50 M, [Y] = 0.10 M, [Z] = 1.0 M [ ] b a d c eq K B A D C = 2.5 5.0 0.40 0.025 50.

[ ] [ ] [ ] Correct Answer: 0.50 K = 0.10 1.0 2.5 5.0 0.40 0.025 50. eq [ ] 2 [ ] 3 0.10 1.0

For the following hypothetical reaction: 2 X(g) + Y(g) A(g) Keq = 250. At a point during the reaction, the concentrations are [A] = 1.0 M, [X] = 0.50 M, and [Y] = 0.10 M. How will the reaction proceed to achieve equilibrium? From left to right (toward products) From right to left (toward reactants) Already at equilibrium

[ ] [ ] [ ] Correct Answer: 1.0 K = 0.50 0.10 From left to right (toward products) From right to left (toward reactants) Already at equilibrium [ ] 1.0 K = [ ] [ ] c 2 0.50 0.10 Because Kc < Keq, the reaction will shift from reactants toward products to achieve equilibrium.

Y(g) + Z(g) 2 A(g) 4 A(g) 2 Y(g) + 2 Z(g) For the following hypothetical reaction: Y(g) + Z(g) 2 A(g) Keq = 4.0  102. Given that, what is the value of Keq for the reaction: 4 A(g) 2 Y(g) + 2 Z(g) 25 5.0 0.016 6.3  102

Correct Answer: 25 5.0 0.016 6.3  102 The second equation is twice the first and reversed. Thus, Keq of the second equation is related to the first as shown below:

The following reaction is at equilibrium: N2(g) + 3 H2(g) 2 NH3(g) If we remove NH3(g), in what direction will the reaction move to reestablish equilibrium? From left to right (toward products) From right to left (toward reactants) No change in equilibrium

Correct Answer: From left to right (toward products) From right to left (toward reactants) No change in equilibrium If [NH3] is decreased, to return to equilibrium, more NH3 must be produced from the reactants. Thus, the reaction shifts from the reactants to the products.

The following reaction is at equilibrium: N2(g) + 3 H2(g) 2 NH3(g) If we increase the volume while holding the temperature constant, in what direction will the reaction move to reestablish equilibrium? From left to right (toward products) From right to left (toward reactants) No change in equilibrium

Correct Answer: From left to right (toward products) From right to left (toward reactants) No change in equilibrium Increasing the volume causes a shift in the equilibrium in the direction that produces more gas molecules, in this case on the reactants side.

The following reaction is at equilibrium: N2(g) + O2(g) 2 NO(g) If we increase the pressure, in what direction will the reaction move to reestablish equilibrium? From left to right (toward products) From right to left (toward reactants) No change in equilibrium

Correct Answer: From left to right (toward products) From right to left (toward reactants) No change in equilibrium The number of gas molecules on both sides of the equation are equal; thus, changing the pressure will not change the equilibrium position.

The following reaction is at equilibrium: N2(g) + O2(g) 2 NO(g) DH° = +180.8 kJ. In what direction will the reaction move to reestablish equilibrium if the temperature is decreased? From left to right (toward products) From right to left (toward reactants) No change in equilibrium

Correct Answer: From left to right (toward products) From right to left (toward reactants) No change in equilibrium As the temperature is lowered, equilibrium will shift to the side of the equation that produces heat. Thus, for an endothermic reaction, the equilibrium shifts from right to left toward the reactants.