Can someone please help? Been stuck for days... In doing the neutralization (usi

Question

Can someone please help? Been stuck for days... In doing the neutralization (using the same calorimeter as in problem 5) of HF with NaOH; 50.0 mL of 2.08M NaOH at 22.2 degree C are mixed with 50.0 mL of 2.08 M HF at 22.2 degree C. After the two solutions were mixed, allow to react, and come to equilibrium temperature; the final temperature was found to be 33.2 degree C. The resulting solution weighed 100.791 g. In order to obtain the specific heat of the resulting NaF solution: 50 mL (51.264 g) of distilled water at 64.3 degree C were added to 52.177 g of the NaF solution from the resulting solution at 22.2 degree C in the same calorimeter. After equilibrium had been reached, the final temperature was found tom be 39.7 degree C. Calculate (a) the specific heat (cal/g- degree C) of the NaF solution, (b) qn (heat of neutralization for the amount used in this experiment) in calories, and (c) Delta Hn (enthalpy of neutralization of HF with NaOH) in the units of (1)cal/mol, (2)kcal/mol, and (3)kJ/mol

Explanation / Answer

Solution :-

50 ml of 2.08 M HF

And 50 ml of 2.08 M NaOH

Initial temperature = 22.0 C

Final temperature = 33.2 C

51.264 g of distilled water at 64.3 C were mixed with 52.177 g NaF solution at 33.2 C

The resulting temperature = 39.7 C

Specific heat of the NaF solution =?

Specific heat of water = 1 cal / g C

a)Water is at higher temperature therefore it looses the heat to raise the temperature of the NaF solution from 33.2 C to 39.7 C

So we can calculate the specific heat of the NaF solution by the following set up

-q hot = q cold

-51.264 g * 1 cal per g C*(39.7 C-64.3 C) = 52.177 g * S * (39.7 C – 33.2 C)

1261.0944 Cal = 339.1505 g. oC *S

1261.0944 Cal / 339.1505 g. oC = S

3.718 cal per g C = S

Therefore the specific heat of the NaF solution is 3.718 cal per g C

b) Now lets calculate the heat of neutralization for the amount of the NaOH and HF used in this experiment

q =m*S*delta T

q= 100.791 g * 3.718 cal per gC* (33.2 C- 22.2 C)

q= 3747 cal

c) Calculating the heat of neutralization per mol of HF and NaOH

moles of HF = 2.08 mol per L * 0.050 L = 0.104 mol

energy given out = 3747 cal

moles = 0.104 mol

enthalpy change per mol = 3747 cal / 0.104 mol

                                             = 36029 cal per mol

So the enthalpy change for the neutralization of the HF and NaOH = -36029 cal per mol

Negative sign because the reaction is exothermic.

Now lets convert cal per mol to Kcal per mol

-36029 cal per mol * 1 Kcal / 1000 cal = -36.029 Kcal /mol

Therefore its -36.029 kcal / mol

Now lets convert it to the kJ/mol

1 kcal = 4.184 kJ

Therefore

-36.029 kcal per mol * 4.184 kJ / 1 cal = -150.7 kJ per mol

Therefore its -150.7 kJ /mol

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