Elementary Principles 405

# Elementary Principles 405 - 9.39 Mass of H 2 SO 4 = Mass of...

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9- 53 9.39 Mass of H SO m1 0 L 1 m o l H S O 1 m L mol H SO 98.02 g 1 mol g H SO Mass of solution m 1 0 L 1 0 mL 1 . 0 6 4 g L 1 mL g solution 24 33 3 3 3 == F H G I K J × 3 3000 2 941 10 3 3192 10 5 6 . . ⇒= × × = × Moles of H O g H O( 1 mol 18.02 g mol H O 22 2 (. . ) ) . 3192 10 2 941 10 161 10 65 5 n mol H O mol H SO mol H O mol H SO mol H O mol H SO 2 2 4 F H G I K J = × = 161 10 3000 536 5 . . ΔΔ Δ ±± ± .. . ,. . HH H f HSO aq .r f o HSO l s HSO aq r 2 2 Table B.1 2 Table B.11 kJ mol kJ mol ej bg 4 4 4 53 6 53 6 81132 7339 884 7 = A = A =+ = = H (3000 mol H SO )(-884.7 kJ / mol H SO ) = -2.65 10 kJ 6 9.40 HCl (aq): Δ ± . H f o f o HCl g s o Tables B.1, B11 kJ mol =+= = 92 31 7514 167 45 NaOH (aq): Δ ± . H f o f o NaOH s s o Tables B1, B.11 kJ mol = = B 4266 4289 469 49 NaCl (aq): Δ ± . H f o f o NaCl s s o Table B.1 Given kJ mol = + = BB 4110 4 87 4061 HCl aq NaOH aq NaCl aq H O l 2 b g +→ + Δ ± . . . H r o kJ mol =− −− 4061 28584 167 45 469 49 550 HCl g NaOH s NaCl s H O l 2 + Δ ± . . . Hv H v H ii r o f o products f o reactants kJ mol kJ mol ∑∑ 4110 28584 92 31 426 6 177 9 The difference between the two calculated values equals
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## This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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