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CHEM 122 Expt #10 Thermochemistry

CHEM 122 Expt #10 Thermochemistry. CCBC-Catonsville. Sample Calculations (p.122). 35.00 mL 1.00 M NaOH. 35.00 mL 1.00 M HCl. T(initial) = 25.2 o C 29.0 o C What is molar H for the reaction?. REACTION: HCl + NaOH HOH + NaCl

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CHEM 122 Expt #10 Thermochemistry

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  1. CHEM 122 Expt #10Thermochemistry CCBC-Catonsville

  2. Sample Calculations (p.122) 35.00 mL 1.00 M NaOH 35.00 mL 1.00 M HCl T(initial) = 25.2oC 29.0 oC What is molar H for the reaction? REACTION: HCl + NaOH HOH + NaCl 1 M 1 M 0.5 M ?? 0.03500 L x 1.00 M HCl = 0.03500 mol HCl Stoichiometry tells us 0.03500 mol NaCl is produced. Molarity of NaCl = 0.03500 mol NaCl/70.00 mL soln = 0.03500 mol NaCl/0.0700 L soln = 0.500 M NaCl

  3. 35.00 mL 1.00 M HCl 35.00 mL 1.00 M NaOH q(rxn) = - q(soln) (not water but 0.500 M NaCl) q = m c T ( or s m T) where c = s = specific heat q(soln) = (mass soln aft mixing)(spec heat of soln)(T of soln) T(soln) = (29.0 – 25.2)oC = 4.2 oC (Watch decimal place!) m (soln) = 70.00 mL (1.03 g/mL) (from table on p. 121)= 72.1 g c (soln) = 4.18 J/(goC) (from table on p.121) q (soln) = (72.1g)(4.18J/goC)(4.2oC) = 1145 J (2 sig. fig.) q (rxn) = - 1145 J (= H because in open container) molar H = - 1145 J / #mol HCl = - 1145 J/0.0350 mol HCl = - 32714 J/mol HCl = -33 kJ/mol HCl T(initial) = 25.2oC 29.0 oC What is molar H for the reaction?

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