Calculations Part 1 REACTION 1 Mass of water 100 g \u0394T Tf Ti 420 C Specific heat

Calculations part 1 reaction 1 mass of water 100 g

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Calculations- Part 1 REACTION 1 Mass of water = 100 g ΔT = Tf – Ti = 4.20 °C Specific heat = 4.18 Joules gramsC qwater (Joules) = 176 0 (3s.f) qwater (kJ) =1.76 (3s.f.) Moles of NaOH = 2.0337 + 1.9658 2 × 39.997 =0.049997 moles qwater (kJ/mol) = 35.2 qreaction (kJ/mol) = -35.2 = ΔH1 REACTION 2 Mass of water = 52g ΔT = Tf – Ti = 7.50°C
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Specific heat = 4.18 Joules gramsC qwater (Joules) =1630 J qwater (kJ) =1.63 J Moles of NaOH = 1.9508 + 1.8342 2 ÷ 39.997 = 0.047316 moles Moles of HCL = 0.05 moles qwater (kJ/mol) = 34.5 qreaction (kJ/mol) = -34.5 = ΔH2 REACTION 3 Mass of water = 100g ΔT = Tf – Ti = 6.65 °C Specific heat = 4.18 Joules gramsC qwater (Joules) = 2780 J qwater (kJ) = 2.78 kJ Moles of NaOH = 0.05 moles Moles of HCL =0.05 moles qwater (kJ/mol) = 60 qreaction (kJ/mol) = -60 = ΔH3 EVALUATING THE VALIDITY OF HESS’S LAW Measured value of ΔH for Reaction 3 = -60 (same as ΔH3 above)
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Value of ΔH for Reaction 3 based on Hess’s law = ΔH2 – ΔH1 = -34.5 +35.2 =0.7 Percent error in the calculated value compared to the measured value: 0.7 60 100 =− 1.2 Also write equations to show how reactions 1 and 2 will be combined to obtain reaction 3: Equations: ( 2 ) NaOH ( s )+ HCl ( aq ) →NaCl ( aq )+ H 2 O ( l ) Δ H =− 95.2 kJ mol ¿ ( 1 ) NaOH ( s ) → NaOH ( aq ) Δ H =− 35.2 kJ mol ____________________________________ __________________ ( 3 ) NaOH ( aq )+ HCl ( aq ) →NaCl ( aq )+ H 2 O ( l ) Δ H =− 60 kJ mol REACTION 4 Mass of water = 100 g ΔT = Tf – Ti = 4.20 °C Specific heat = 4.18 Joules gramsC qwater (Joules) = 176 0 (3s.f) qwater (kJ) =1.76 (3s.f.) Moles of NaOH = 2.0337 + 1.9658 2 × 39.997 =0.049997 moles
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qwater (kJ/mol) = 35.2 qreaction (kJ/mol) = -35.2
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