Hello there, I am a little confused with the results I got after doing this coffee cup calorimetry experiment. If you could please review my formulas to see if I am doing everything properly I would be most appreciative.
For a reaction between baking soda and vinegar:
10.06 g of vinegar were used as well as 2.45 g of baking soda. Total mass is 12.51 g.
Temp(i): 17.2˚C
Temp(f): 16.6˚C
The ∆Hrxn, enthalpy of reaction was calculated by:
12.51 g * 4.148 J/g ˚C * (16.6˚C - 17.2˚C) = -31.41 J
-∆H = 31.41 J or 0.031 KJ This is an endothermic reaction
Now for covering the ∆Hrxn to a molar value:
10.06 g (1/60.05) = 0.17 mol
2.45 g (1/84.00) = 0.030 mol
Total molar mass = 0.20 mol
Then do ∆Hrxn / n
0.031 KJ / 0.20 mol = 0.155 KJ/mol
Again, if you could please tell me if I made any errors I would be very thankful.
Hello! Let's review your calculations for the coffee cup calorimetry experiment.
First, let's calculate the heat change (∆H) for the reaction using the given data:
Mass of the system (baking soda + vinegar) = 12.51 g
Specific heat capacity of the system = 4.148 J/g˚C
Change in temperature (∆T) = 16.6˚C - 17.2˚C = -0.6˚C
∆H = Mass * Specific heat capacity * ∆T
∆H = 12.51 g * 4.148 J/g ˚C * (-0.6 ˚C)
∆H = -31.41 J
It seems like you have correctly calculated the heat change (∆H) as -31.41 J. The negative sign indicates that the reaction is exothermic, not endothermic as mentioned in your description.
To convert the ∆H to kilojoules, you can divide by 1000:
∆H = -31.41 J = -0.03141 kJ (rounded to five decimal places)
Now, let's calculate the molar ∆H (∆Hrxn) using the number of moles of reactants used in the experiment.
Number of moles of vinegar (CH3COOH):
10.06 g * (1 mol / 60.05 g) = 0.1675 mol (rounded to four decimal places)
Number of moles of baking soda (NaHCO3):
2.45 g * (1 mol / 84.00 g) = 0.0292 mol (rounded to four decimal places)
The total moles of reactants used in the experiment is:
0.1675 mol + 0.0292 mol = 0.1967 mol (rounded to four decimal places)
Now, we can calculate the molar ∆H (∆Hrxn) by dividing the heat change (∆H) by the total moles:
∆Hrxn = ∆H / n
∆Hrxn = -0.03141 kJ / 0.1967 mol
∆Hrxn = -0.15958 kJ/mol (rounded to five decimal places)
It seems like you have made a small rounding error when calculating the molar ∆H (∆Hrxn). The correct answer is approximately -0.15958 kJ/mol, not 0.155 kJ/mol.
Overall, your calculations seem to be mostly accurate, with the exception of the sign of ∆H and a rounding error when calculating the molar ∆H (∆Hrxn). I hope this helps straighten things out for you! Let me know if you have any further questions.