Calculate the delta H degree (on a per mole basis) for the HCl/N_aOH neutralizat

Question

Calculate the delta H degree (on a per mole basis) for the HCl/N_aOH neutralization. You may assume that the specific heat and density of the resulting solution to be the same as water. Be sure to use the combined total volumes of HCl and N_aOH n calculating the total mass for this reaction. Remember to take into account the moles of limiting reagent when calculating the delta H degree on a per mole basis Calculate the average value for the molar heat of neutralization for your two trials. Given that the literature value for the heat of neutralization of h H_3O^+ is - 56.146 kJ/mol, Experiment details: Polystyrene drinking cups are placed inside of each other to serve as the calorimeter Inner cup will be covered with a lid to minimize heat gain/loss during chemical reaction. I know this is a big question but I am beyond lost, Please help me. It would be very appreciated!

Explanation / Answer

Calorimetry

a. Heat q = mCpdT

with

m = total mass of solution

Cp = specific heat of water

dT = change in temperature

Thus,

With data set 1

total mass of solution (m) = 50 + 38 = 88 ml

                                         = 88 ml x 1 g/ml

                                         = 88 g

density of water = 1 g/ml

change in temperature (dT) = 5.8 oC

So,

heat released = 88 x 4.18 x 5.8/1000 = 2.133 kJ

Heat of neutralization per mole of NaOH = 2.133 kJ/0.9817 M x 0.038 L

                                                                 = 57.18 kJ/mol

With data set 2

total mass of solution (m) = 50 + 38 = 88 ml

                                         = 88 ml x 1 g/ml

                                         = 88 g

density of water = 1 g/ml

change in temperature (dT) = 4.9 oC

So,

heat released = 88 x 4.18 x 4.9/1000 = 1.802 kJ

Heat of neutralization per mole of NaOH = -1.802 kJ/0.9817 M x 0.038 L

                                                                 = -48.31 kJ/mol

Average heat of netralization = (-57.18 - 48.31)/2

                                               = -52.74 kJ/mol

b. So the literature value is higher than the value calculated by this experiment. The error could be due to incorrect measurement of reaction volume or temperature change.

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