chemistry IA Urgent help

[meenaa]

HI,

I need urgent help with chemistry IA. My IA is on Calorimetric Determination of the Enthalpy Change Using Hess’ Law. I am having trouble with calculation and trying to calculate with uncertainties. Here is my results

RESULTS/DATA

1. NaHCO₃

Time

5 sec

10 sec

15 sec

20 sec

25 sec

30 sec

Trial One

14

13

13

12

13

11

Weighing boat before = 2.249g

Weighing boat with contents= 7.001g

After = 2.354g

Temp of solution= 13^oC

Trial Two

20

17

15

14

14.5

13

Weighing boat before =2.255g

Weighing boat with contents=7.002g

After= 2.443g

Temp of solution =21^oC

Trial Three

18

15

14.5

13

13.5

12

Weighing boat before =2.261g

Weighing boat with contents=7.001g

After= 2.423g

Temp of solution=21^oC

Trial Four

17

15.5

14.5

14

13.5

13

Weighing boat before =2.334g

Weighing boat with contents= 7.001g

After = 2.339g

Temp of solution=18.5^oC

2. Na₂CO₃

Time

5 sec

10 sec

15 sec

20 sec

25 sec

30 sec

Trial One

18

19

20

21

21.5

22

Weighing boat before =2.328g

Weighing boat with contents=4.001g

After = 2.337g

Temp of solution= 17^oC

Trial Two

19

19.5

20

21

21.5

21.5

Weighing boat before =2.345g

Weighing boat with contents=4.001g

After = 2.188g

Temp of solution=17^oC

Trial Three

20.5

21

22

23

23.5

23.5

Weighing boat before = 2.324g

Weighing boat with contents=4.001g

After = 2.333g

Temp of solution=19^oC

Trial Four

19.5

20

20.5

21

21.5

21.5

Weighing boat before = 2.324g

Weighing boat with contents=4.001g

After = 2.344g

Temp of solution=18^oC

Uncertainty

· Measuring cylinder: 1.2 +/-

· Thermometer: 0.25 +/-

· Electrical balance: 0.001 +/-

this is calculation that i did

ASSUMPTIONS

The solutions made are close enough to being water (but are not water) that their specific heats are also 4.18 just like water.

Since the density of water is 1.00g/mL, and the solutions formed are close enough to be like water, their density will also be the same as that of water

ANALYSIS AND DISCUSSION

Sodium Hydrogen Carbonate decomposes upon heating to give sodium carbonate, water and Carbon dioxide gas as illustrated by the equation below:

2NaHCO_3(s) ---------------> Na₂CO_3(s) + H₂O_(l) + CO_2(g) = ∆H₁

∆H₁ cannot, however, be directly calculated but can be found using enthalpy changes from two other reactions of Hydrochloric Acid with Sodium Hydrogen Carbonate and Sodium Carbonate respectively.

NaHCO_3(s) + 2HCl_(aq) ------------>NaCl_(aq) + 2H₂O_(l) + 2CO_2(g) = ∆H₂

Average Change in Temperature = {(13-14)+(21-20)+(21-18)+(18.5 -17)}÷4

= 1.125^oC ≈ 274.275K

Average Mass = {(7.001-2.249)+(7.002-2.255)+(7.001-2.261)+(7.001-2.334)}÷4

= 4.727g

Energy Balance: q_rxn = q_sln + q_cal

Where:

q_rxn – The Energy produced/ expended in the reaction

q_sln– The energy content of the solution

q_cal– The energy content of the calorimeter

Thus;

Q_rxn = Specific Heat Capacity × Mass of the solution × Temperature Change

= 4.18 × 4.727 ×274.275

∆H₂ = 5419.36 J/mol

Na₂CO_3(s) + 2HCl_(aq) ----------------------> 2NaCl_(aq) + CO_2(g) + H₂O_(l)

Average Change in Temperature = {(17-18)+(17-19)+(19-20.5)+(18-19.5)}÷4

= -1.5^oC ≈ 271.65K

Average Mass = {(4.001-2.328)+(4.001-2.345)+(4.001-2.324)+(4.001-2.324)} ÷4

= 1.84g

Energy balance: q_rxn + q_sln +q_cal = 0

Therefore, q_rxn = - ( q_sln+ q_cal)

Q_rxn = Specific Heat Capacity × Mass of the solution × Temperature Change

= - (4.18 × 1.84 × 271.65)

= - 2089.314 J/mol

Therefore;

∆H₂ = 5419.36 J/mol

∆H₃ = -2089.314 J/mol

Thus;

∆H₁ = ∆H₂ + ∆H₃

∆H₁ = 5419.36 + (-2089.314) =>

∆H₁ = 3330.046 J/mol

My friends have results in range of 90. I am not sure where i have gone wrong. And can someone help me with how to calculate using the uncertainty figures in the calculation.

I have done my calculation by averaging out the total results for one solutions. I am i doing it wrong or should i be calculation hews law individually and then averaging it out.

can someone please help me.

thankyou in advance

meena

[twilight]

I think you should be using the graphical method to determine the temperature change (the one where you plot a graph of temperature against time then extrapolate to determine the lowest temperature) for trial 1, 2, 3 and 4 each, then average the 4 to get your temperature change for each solution.