You want to determine H o for the reaction Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) To

Question

You want to determine H o 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 H o is known: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) H o = 57.32 kJ

(a) Calculate the heat capacity of the 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: 16.9°C Maximum T recorded during reaction: 30.4°C Density of resulting NaCl solution: 1.04 g/mL c of 1.00 M NaCl(aq): 3.93 J/g·K

kJ/°C

(b) Use the result from part (a) and the following data to determine H o rxn for the reaction between zinc and HCl(aq): Amounts used: 100.0 mL of 1.00 M HCl and 1.3078 g of Zn Initial T of HCl solution and Zn: 16.8°C Maximum T recorded during reaction: 20.5°C Density of 1.0 M HCl solution: 1.015 g/mL c of resulting ZnCl2(aq): 3.95 J/g·K

kJ/mol

Explanation / Answer

a)

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*(20.5-16.8) + 0.1*HRxn + 101.5*3.95*(20.5-16.8) = 0

HRxn =( -0.01587*(20.5-16.8) +  101.5*3.95*(20.5-16.8) ) /0.1

= 14834.2837J/mol

= 14.83 kJ/mol

the basis is stated per unit mol of HCl

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