Question

Please use part C data to solve question 1. The theoretical value is -57.3kJ/mol to compare answer to. Calorimeter constant is 22.1J/gC

PART C Data Analysis: 1. Calculate the heat of neutralization (AH.eu) for each trial 2. Calculate the average, standard deviation, and relative percent standard deviation for your three trials 3. Calculate the percent error based on your average and the theoretical value given in your data set. Data Trial 3 45.120 g 105.687 g Trial 1 Trial 2 Part A (calorimeter constant = 22.1 J/°C) Metal name lead Mass of calorimeter 44.193 g 47.272 g Mass of calorimeter + water 115.543 g | 125.569 g Mass of lead 98.059 g Initial temperature lead 100.0 °C 100.0 °C Initial temperature of water 21.0°C 21.6°C Final temperature 24.0 °C 24.2 °C Part B Mass of calorimeter 41.304 g 50.267 g Mass of calorimeter + water 65.390 g 73.507 g mass of cesium hydroxide 3.120 g 5.199 g Initial temperature water/calorimeter 24.8 °C 22.0°C Final temperature 36.4 °C 42.2 °C 100.0 °C 23.0 °C 26.0 °C Trial 1 Trial 2 Trial 3 Part C Volume NaOH 21.2 mL 23.0 mL 20.2 mL Volume HCI 20.4 mL 20.4 mL 22.2 mL Initial temperature solution 23.0°C 20.6°C 23.6 °C Final temperature 40.0 °C 37.2 °C 41.4 °C Part D Mass of calorimeter 49.441 g 48.532 g 49.058 g IMass of calorimeter + water 100.671 g 102.874 g | 96.915 g Initial temperature water 23.2 °C 24.4 °C 25.4 °C Mass of calorimeter + water + ice 107.485 g 110.007 g | 104.923 g Final temperature 11.8 °C 13.0°C 12.0°C Theoretical values for data analysis Part A = 0.128 J/gºC Part B = -71.6 kJ/mol Part C = -57.3 kJ/mol Part D = 334 J/g 46.949 g 74.224 g 5.527 g 21.6 °C 41.4 °C

Answer 1

Answer:

Using Part C data to estimate the heat of neutralization ∆Hneu for the reaction between an acid (HCl) and a base(NaOH).

So, the neutralization reaction is

NaOH + HCl$\rightarrow$ NaCl + H2O. ∆Hneu = -? kJ/mol...(1)

So, from above equation we can say that heat of neutralization is the amount of energy released for formation of one mole of water.

H+ (aq) + OH-(aq) $\rightarrow$ H2O(aq) ...........(2)

This amount of energy released can be easily calculated by applying laws of Thermodynamics(Law of conservation of energy) as follows:

Here System is reaction as per eq(2) above.

Surroundings is solution+calorimeter

So, heat released by reaction = heat absorbed by solution+ heat gained by calorimeter

-q(neu) = q(solution)+q(calorimeter)............(3)

We know that, q= mc∆T

Where, m = mass

c = specific heat capacity

∆T= change in temp.

As the heat of neutralization is expressed in kilo joule per mole, we need to exactly find the number of moles of reactants present in the solution.

I. '' Estimating limiting reagent''

the reactant with less number of moles is limiting reagent

For trial 1 , Vol. Of NaOH= 21.2ml, Molarity given as 3M

Molarity = moles /volume in lit

Or 3 = moles/0.0212 lit

Moles of NaOH = 0.0636

Similarly, for HCl, V= 20.4 ml , Molarity= 3M

Moles of HCl= 0.0612

So, our limiting reagent isHCl with0.0612 moles

Follow steps to find ∆Hneu

Step 1: Calculating q(solution)

Mass of solution= density of mixture * total volume of mixture

=1 g/ml * 41.6 ml (given that density of solution is 1 gram per ml)

= 41.6 g

q(sol) = mc(40-23)

= 41.6 g • 4.184 J/g-°C • 17 °C

= 2958.92 J

Step 2: calculating q calorimeter

q (calorimeter) = Ccal•∆T

= 22.1 J/°C •(40-17)°C ( assuming same ∆T as of solution(

= 375.7 J

So, as per eq 3

-qneu = 2958.92 + 375.7

Or qneu =- 3334.62 J

Step 3: estimating heat of neutralization

Heat of neutralization is: qneu /no. Of moles of Limiting reagent

Or ∆H neu = -3334.62/0.0612

= -54487.25 J/mol

= - 54.487 kJ/mol.

Redo the above calculations For Trial 2

a. Here limiting reagent is again HCl

No. Of moles of HCl = 3* (20.4/1000)

= 0.0612 moles

b. Estimating q sol

Total Vol of sollution = 23+20.4 = 43.4 mL

Mass = 1 g/mL • 43.4 g

= 43.4g

q sol = mc∆T

= 43.4* 4.184 *(37.2-20.6)

= 3014.32 J

c. Estimating q calorimeter

= Ccalorimeter •∆T

= 22.1 (37.2-20.6)

= 366.86

d. Calculating heat of neutralization

-q neu =3014.32 + 366.86

Or qneu = - 3381.18 J

Or ∆H neu = qneu /numberof moles of limiting reagent

=-3381.18/0.0612

= -55.248 kJ/mol

For trial third

a.Herethe limiting reagent is NaOH

No. Of moles = 3 *(20.2 /1000) = 0.0606 moles

b. Estimating q solution

Total volume is 42.4 ml

So, mass = 42.4 g

qsol = mc∆T

= 42.4*4.184*(41.4- 23.6)

= 3157.75 J

c. Estimating q calorimeter

= C∆T

=22.1(41.4-23.6)

= 393.38 J

d. Estimating heat of neutralization

So, - qnue =3157.75 + 393.38

Or q neu=- 3551.13 J

Or ∆H neu = -3551.13/ 0.0606

= -58599.5 J/mol

∆H neu=-58.59 kJ/mol