value 10.00 points Be sure to answer all parts. You want to determine AH for the

Question

value 10.00 points 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 foll reaction, whose AH is known: AH 57.32 kJ (a) Calculate the heat capacity ofthe calorimeter from these data: Amounts used: 50.0 mL of 2.00 M HCl and 50.0 mL of 2.00 M NaoH Initial T of both solutions: 16.9oC Maximum T recorded during reaction: 30.4oC Density of resulting Naci solution: 1.04 g/mL c of 1.00 M Naci(aq): 3.93 J/g K kJPC References eBook & Resources

Explanation / Answer

Q1

For heat capacity:

Qcalorimeter + Qreaction + Qwater = 0

Qcal = C*(Tf-Ti)

Qreaction = n*HRxn

Qwater = m*Cw*(Tf-Ti)

so

mol of reaction used = MV = 50*2 = 100 mmol = 0.1 mol

Qreaction = -0.1*57.32 = -5.732 kJ = -5732 J

m water = D*V = (1.04)(50+50) = 104 g

Qwater = 104*3.93*(30.4-16.9) = 5517.72 J

so..

Ccal*(30.4-16.9) - 5732 + 5517.72= 0

Ccal = (5732 - 5517.72) /(30.4-16.9) = 15.87 J/°C = 0.015.87 kJ/°C

b)

now use this for Zn reaction:

Qcalorimeter + Qreaction + Qwater = 0

n = MV = 100 mmol of HCl = 0.1 mol

mol of Zn = mass/MW = 1.3078/65.38 = 0.02 mol

there is HCl limiting, so 0.1 mol of

mass of solution = D*V = (100*1.015) = 101.5 g

0.01587*(21.1-16.8) + 0.1*HRxn + 101.5*3.95*(21.1-16.8) = 0

HRxn =( -0.01587*(21.1-16.8) +  101.5*3.95*(21.1-16.8) ) /0.1 = 17239.09259J/mol = 17.23 kJ/mol

the basis is stated per unit mol of HCl

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