Please show all calculations. Please show all calculations. B. Determination of

Question

Please show all calculations. Please show all calculations.

B. Determination of the Enthalpy of Neutraliation HCI NaOH Molarity of acid and base, M Volume of acid and base, mL Volume of acid and base, L mL HCI NaOH Temperature Readings, °C Initial (0) 1 minute 2 minutes 2030 2X220.yo 2011 minutes 20 50 4 minutes Pour NaOH into HCI 266 5 minutes 6 minutes 7 minutes 8 minutes 9Se 26.50 26.90 26.90 Initial temperature (from graph) Average initial temperature Final temperature (from graph), Change in temperature, AT Energy absorbed by surroundings, Qabs Energy released by reaction, grel Limiting Reagent Moles of limiting reagent Enthalpy of neutralization, AH oc °C oc & SaZ moles J/mole

Explanation / Answer

Enthalpy of neutralization

Total volume = 100 ml

density of water = 1 g/ml

total mass = 100 ml x 1 g/ml = 100g

specific heat of water = 4.184 J/g.oC

Average temperature (acid + base) initial = 20.5 oC

Average final temperature = 26.7 oC

Change in temperature = 6.2 oC

Heat released by reaction = 100 x 4.184 x 6.2 = -2594.1 J

Heat absorbed by surrounding = 2594.1 J

limiting reactant = HCl

moles of HCl = 1.014 M x 0.05 L = 0.0507 moles

Enthalpy of neutralization dH = -2594.1 J/0.0507 moles x 1000 = 51.16 kJ/mol

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