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Suppose you had used 4.00 grams of NaOH(s) in reaction 1. What would have been the number of Joules released in the reactions? What effect would this have on the value of X1?

Suppose you had used 4.00 grams of NaOH(s) in reaction 1. What would have been the number of Joules released in the reactions? What effect would this have on the value of X1?

Chemistry for Engineering Students
Chemistry for Engineering Students
3rd Edition
ISBN:9781285199023
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter1: Introduction To Chemistry
Section: Chapter Questions
Problem 1.77PAE
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Suppose you had used 4.00 grams of NaOH(s) in reaction 1. What would have been the number of Joules released in the reactions? What effect would this have on the value of X1?

Results Data Reaction Mass of NaOH Volume of NaOH + HCI or H20 Mass of solution Initial Temperature Final Temperature DT 3 1 1.00 g 50.0 mL 2 1.00 g 50.0 mL 25.0 mL 51.0 g 25.00 C 30.30 C 30.30 – 25.00 = 5.30 °C 51.0 g 25.00 C 36.97 C 36.97 – 25.0 = 11.97 C 51.0 g * 4.18 * 5.30 | 51.0 g * 4.18 * 50.0 g 25.00 C 31.7 С 31.7 – 25.00 = 6.7 •C 50.0 g * 4.18 * 6.7 = -1400.3 j/-1.4 kJ - %3D %D Heat of reaction = 1129 j/ 1.129 kJ 11.97 = -2500 j/-2.5 kJ 0.0250 mol -2.5/0.025 = -100.0 kJ/mol Mol NaOH Enthalpy, kJ/mol 0.025 mol 1.129/0.025 = -45 kJ/mol 0.025 mol -1.4/0.025 = -56.0 kJ/mol 1. Net lonic Equation Reaction 1: NaOH(s) → Na* (aq) + OH" (aq) X1 AH(endothermic)= 1129 J/ 1.129 kJ AH(heat evolved)= -45 kJ/mol Reaction 2: Na* (aq) + OH (aq) + H* (aq) + Cl (aq) → H20 (1) + Na* (aq) + CI (aq) X2 AH(exothermic)= -2500 J/-2.5 kJ AH(heat evolved)= -100.0 kJ/mol Reaction 3: NaOH (s) + H* (aq) + CI- (aq) → H20 (1) + Na* (aq) + Cl (aq) X3 AH(exothermic)= -1400.3 J/-1.4 kJ AH(heat evolved)= -56.0 kJ/mol
Transcribed Image Text:Results Data Reaction Mass of NaOH Volume of NaOH + HCI or H20 Mass of solution Initial Temperature Final Temperature DT 3 1 1.00 g 50.0 mL 2 1.00 g 50.0 mL 25.0 mL 51.0 g 25.00 C 30.30 C 30.30 – 25.00 = 5.30 °C 51.0 g 25.00 C 36.97 C 36.97 – 25.0 = 11.97 C 51.0 g * 4.18 * 5.30 | 51.0 g * 4.18 * 50.0 g 25.00 C 31.7 С 31.7 – 25.00 = 6.7 •C 50.0 g * 4.18 * 6.7 = -1400.3 j/-1.4 kJ - %3D %D Heat of reaction = 1129 j/ 1.129 kJ 11.97 = -2500 j/-2.5 kJ 0.0250 mol -2.5/0.025 = -100.0 kJ/mol Mol NaOH Enthalpy, kJ/mol 0.025 mol 1.129/0.025 = -45 kJ/mol 0.025 mol -1.4/0.025 = -56.0 kJ/mol 1. Net lonic Equation Reaction 1: NaOH(s) → Na* (aq) + OH" (aq) X1 AH(endothermic)= 1129 J/ 1.129 kJ AH(heat evolved)= -45 kJ/mol Reaction 2: Na* (aq) + OH (aq) + H* (aq) + Cl (aq) → H20 (1) + Na* (aq) + CI (aq) X2 AH(exothermic)= -2500 J/-2.5 kJ AH(heat evolved)= -100.0 kJ/mol Reaction 3: NaOH (s) + H* (aq) + CI- (aq) → H20 (1) + Na* (aq) + Cl (aq) X3 AH(exothermic)= -1400.3 J/-1.4 kJ AH(heat evolved)= -56.0 kJ/mol
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