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Chapter 20 Pract - -167158 =-8.314 698.15 ln(K K = 3.22 10 12 6 Solve for the Q of the reaction if there is 3.250 atm of Cl and 0.050 atm of Cl 2

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Chapter 20 Practice Given the chemical reaction: 2 Cl ( g ) Cl 2 ( g ) And the following information: ΔH f (kJ/mol) S 0 (J/mol-K) Cl 2 ( g ) 0 223.0 Cl ( g ) 121.0 165.1 1. Solve for the ΔH of the reaction. ΔH = 0 – 2 * 121 = -242 kJ/mol 2. Solve for the ΔS of the reaction ΔS = 223.0 – 2 * 165.1 = -107.2 J/mol-K 3. Solve for ΔG 0 of the reaction at 425 Celsius. ΔG 0 = -242000 – 698.15 * -107.2 = -167158 J/mol = -167 kJ/mol 4. Is the reaction spontaneous at 425 Celsius with the standard concentrations? Yes 5. Solve for the K of the reaction at 425 Celisus.
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Unformatted text preview: -167158 = -8.314 * 698.15 * ln(K) K = 3.22 * 10 12 6. Solve for the Q of the reaction if there is 3.250 atm of Cl and 0.050 atm of Cl 2 at the same temperature. Q = (0.05/(0.08206 * 698.15)) /( 3.25/(0.08206 * 698.15)) 2 = 0.2712 7. Solve for ΔG of the reaction with the concentrations in question 6 at 425 Celsius ΔG = -167,158 + 8.314 * 698.15 * ln 0.2712 = -174732 J/mol 8. Is the reaction spontaneous if there is 3.250 atm of Cl and 0.050 atm of Cl 2 ? Yes...
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This note was uploaded on 11/29/2011 for the course CHEM 1212 taught by Professor Dockery during the Fall '08 term at Kennesaw.

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