saved 1 attempts left Check my work Be sure to answer all parts. You want to det
ID: 713064 • Letter: S
Question
saved 1 attempts left Check my work Be sure to answer all parts. You want to determine AH for the reaction Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) To do so, you first determine the heat capacity of a calorimeter using the following reaction, whose AT 0 is known: NaOH(aq) + HCl(aq) NaCl(aq) + H20() AH57.32 kJ (a) Calculate the heat capacity of the calorimeter from these data: Amounts used: 50.0 mL of 2.00 MHCI and 50.0 mL of 2.00 M NaOH Initial T of both solutions: 16.9°C Maximum T recorded during reaction: 30.4°C Density of resulting NaCl solution: 1.04 g/mL c of 1.00 M NaCl(a): 3.93 J/g K (b) Use the result from part (a) and the following data to determine AHfor the reaction between zinc and HC(aq): rxn Amounts used: 100.0 mL of 1.00 M HCI and 1.3078 g of Zn Initial T of HCl solution and Zn: 16.S°C Maximum T recorded during reaction: 20.3°C Density of 1.0 M HCI solution: 1.015 g/mL c of resulting ZnCl (ag): 3.95 J/g K Report Hi k.J/molExplanation / Answer
Ans. Part A. Step 1: Moles of HCl taken = Molarity x Volume of solution in liters
= 2.0 M x 0.050 L
= 0.10 mol
Moles of NOH taken = 2.0 M x 0.050 L = 0.10 mol
# According to the stoichiometry of balanced reaction, 1 mol NaOH reacts with 1 mol HCl to form 1 mol H2O of neutralization.
In the given reaction mixture, the moles of NaOH is equal to that of HCl.
So, moles of H2O formed = 0.10 mol
# Total heat released during neutralization = dH0 x moles of H2O formed
= (-57.32 kJ / mol) x 0.10 mol
= -5.732 kJ
= -5732.0 J
# Step 2: Volume of reaction mixture = 50.0 mL (HCl) + 50.0 mL (NaOH) = 100.0 mL
Mass of solution = Volume x Density = 100.0 mL x (1.04 g/ mL) = 104.0 g
# Step 3: Some of the heat (from total of 5732.0 J) is absorbed by solution while the rest is taken up by calorimeter.
The amount of heat gained by solution during increase in temperature is given by the equation-
q = m s dT - equation 1
Where,
q = heat gained
m = mass
s = specific heat
dT = Final temperature – Initial temperature
Putting the values in equation 1-
q = 104.0 g x (3.93 kJ g-10C-1) x (30.4 – 16.9)0C
Hence, q = 5517.72 J
# Therefore, heat gained by the solution = 5517.72J
# Now,
Heat gained by calorimeter =
Total heat released during neutralization – heat gained by solution
= 5732.0 J - 5517.72 J
= 214.28 J
And,
Calorimetric constant = Heat gained by calorimeter / Increase in temperature
= 214.28 J / (13.50C)
= 15.873 J 0C-1
= 0.015873 kJ 0C-1
# Part B: Step 1:
Mass of solution = Volume x Density = 100.0 mL x (1.015 g/ mL) = 101.5 g
Increase in temperature, dT = (20.3- 16.8)0C = 3.50C
The total heat released during reaction of Zn with HCl must be equal to the total heat gained by solution and calorimeter-
Or, Q = m s dT (solution) + C dT (calorimeter)
Or, Q = (101.5 g x 3.95 J g 0C-1 x 3.50C) + (15.873 J 0C-1 x 3.50C)
Or, Q = 1403.2375 J + 55.5555 J
Hence, Q = 1458.793 J
Therefore, total heat released during reaction of Zn with HCl = 1458.793 J
# Step 2: Moles of Zn taken = Mass / MW = 1.3078 g/ (65.39 g/ mol) = 0.02 mol
Now (assuming Zn to be the limiting reactant)-
dH0rxn = Heat released during reaction/ Moles of Zn
= 1458.793 J / 0.02 mol
= 72939.65 J/ mol
= 72.94 kJ/ mol
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.