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?

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ISBN:9781305957404
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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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