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6. For question 6, recall that the value of K will change in the following ways if an equilibrium equation is changed: Writing the equation in reverse will invert the equilibrium expression. Mathematically, if you already have the K value, then this is the same as K,everse= • Multiplying the coefficients by a common factor will raise K to the corresponding power. This means if you multiply an equation by 2, the equilibrium expression is raised to the 2nd power. If you already have the K value, then K is then equal to K’. a. If the value of K is 2.8 x 10° at 1000 K for the following reaction, what is the K value for the reverse reaction? 250, (8) + O; (8) → 2 SO, (8) b. If the K value for 250; (g) + O; (g) → 2 so, (g) is 2.8 x 10°, what is the K value for the following reaction? 4 SO (g) + 2 0; (g) 4 SO, (g) c. Calculate the value of K for the reaction: Caco, (s) + H" (ag) → Ca² (ag) + HCO," (ag) Given the following equilibrium constants: Caco, (s) + Ca?" (ag) + CO, (ag) K, = 1043 HCO, (ag) + H" (ag) + CO, (ag) K2 = 1010 %3D |

6. For question 6, recall that the value of K will change in the following ways if an equilibrium equation is changed: Writing the equation in reverse will invert the equilibrium expression. Mathematically, if you already have the K value, then this is the same as K,everse= • Multiplying the coefficients by a common factor will raise K to the corresponding power. This means if you multiply an equation by 2, the equilibrium expression is raised to the 2nd power. If you already have the K value, then K is then equal to K’. a. If the value of K is 2.8 x 10° at 1000 K for the following reaction, what is the K value for the reverse reaction? 250, (8) + O; (8) → 2 SO, (8) b. If the K value for 250; (g) + O; (g) → 2 so, (g) is 2.8 x 10°, what is the K value for the following reaction? 4 SO (g) + 2 0; (g) 4 SO, (g) c. Calculate the value of K for the reaction: Caco, (s) + H" (ag) → Ca² (ag) + HCO," (ag) Given the following equilibrium constants: Caco, (s) + Ca?" (ag) + CO, (ag) K, = 1043 HCO, (ag) + H" (ag) + CO, (ag) K2 = 1010 %3D |

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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6. For question 6, recall that the value of K will change in the following ways if an equilibrium equation is changed: Writing the equation in reverse will invert the equilibrium expression. Mathematically, if you already have the K value, then this is the same as Kreverse= Multiplying the coefficients by a common factor will raise K to the corresponding power. you multiply an equation by 2, the equilibrium expression is raised to the This means 2nd power. If you already have the K value, then K is then equal to K?. a. If the value of K is 2.8 x 10? at 1000 K for the following reaction, what is the K value for the reverse reaction? 250, (g) + 02 (g) → 2 SO; (g) b. If the K value for 250, (g) +02 (g) <> 2 SO, (g) is 2.8 x 10?, what is the K value for the following reaction? 4 SO, (g) + 2 02 (g) > 4 SO; (g) Calculate the value of K for the reaction: Caco, (s) + H* (ag) Ca?- (ag) + HCo, (ag) Given the following equilibrium constants: Caco; (s) > Ca²+ (ag) + CO,? (ag) K = 1063 HCO, (ag) > H* (ag) + Co,? (ag) K2 = 10103 |
Transcribed Image Text:6. For question 6, recall that the value of K will change in the following ways if an equilibrium equation is changed: Writing the equation in reverse will invert the equilibrium expression. Mathematically, if you already have the K value, then this is the same as Kreverse= Multiplying the coefficients by a common factor will raise K to the corresponding power. you multiply an equation by 2, the equilibrium expression is raised to the This means 2nd power. If you already have the K value, then K is then equal to K?. a. If the value of K is 2.8 x 10? at 1000 K for the following reaction, what is the K value for the reverse reaction? 250, (g) + 02 (g) → 2 SO; (g) b. If the K value for 250, (g) +02 (g) <> 2 SO, (g) is 2.8 x 10?, what is the K value for the following reaction? 4 SO, (g) + 2 02 (g) > 4 SO; (g) Calculate the value of K for the reaction: Caco, (s) + H* (ag) Ca?- (ag) + HCo, (ag) Given the following equilibrium constants: Caco; (s) > Ca²+ (ag) + CO,? (ag) K = 1063 HCO, (ag) > H* (ag) + Co,? (ag) K2 = 10103 |
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