Name, Lab Section, Date Data Table -Enthalpy of Neutralization 1. Mass of Cups S

Question

Name, Lab Section, Date Data Table -Enthalpy of Neutralization 1. Mass of Cups Stir Bar to 3 decimal places 10. 218 2. Mass of Cups+Stir Bar + Reaction Mixture to 3 decimal places IOD.GTI 30·93 23.93 SM of Rton Mitur to 3 decimal l 4. Final Temperature to 2 decimal places: T 5. Initial Temperature to 2 decimal places: Tr 6. Change in Temperature: 30. 393 23.927 °C , oC Calculate the heat of neutralization in kJ/mole using the following formula: kJ mel Heat of Reaction . -an-q· 10001 Where: m solution mass (3 above spesific heat capacty of 1 M salt water-3.89 /e C temperature change(6 above) . Tn. Tilib n = moles HaO formed mole Write the net ionic equation for this acid-base reaction. Cay) Con J Rev D-8-2017 Page 9 of 10

Explanation / Answer

Heat of neutralization

from the experiment

dHneutralization = -msdT/moles

m = 100.571 g

s = 3.89 J/g.oC

dT = 6.4 oC

moles = 0.05 mol

So,

dHneutralization = -100.571 x 3.89 x 6.40/0.05 x 1000 = -50.10 kJ/mol

Net ionic equation for neutralization,

H+(aq) + OH-(aq) <==> H2O(l)

From standard enthalpy data,

dHrxn = dH(products) - dH(reactants)

           = -285.8 - (-230) = -55.8 kJ/mol

%error = (-50.1 - (-55.8)) x 100/-55.8 = 10.21%

1. molarity of acid or base needed to get a temperature change of 40 oC

moles = -100.571 x 3.89 x 40/-50.10 x 1000 = 0.312 mol

molarity needed = 0.312 mol/0.05 L = 6.24 M

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