You want to determine H° for the reaction Zn(s) + 2HCl(aq) ZnCH(aq) + H2(g) (a)

Question

You want to determine H° for the reaction Zn(s) + 2HCl(aq) ZnCH(aq) + H2(g) (a) To do so, you first determine the heat capacity of a calorimeter using the following reaction, whose H is known: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) Ar = _ 57.32 kJ Calculate the heat capacity of the calorimeter from these data: Amounts used: 50.0 mL of 2.00 M HCI and 50.0 mL of 2.00 M NaOH Initial T of both solutions: 16.9°C Maximum Trecorded during reaction: 30.4°C Density of resulting NaCl solution: 1.04 g/mL C of 1.00 M NaCl(aq) = 3.93 J/gK

Explanation / Answer

No. of moles of NaOH taken = Molarity*Volume = 2*50/1000 = 0.1 mole

No. of moles of HCl taken = Molarity*Volume = 2*50/1000 = 0.1 mole

One mole NaOH gives one mole NaCl and one mole water,

Therefore, No. of moles of NaCl formed in the reaction = 0.1 mole

Therefore, No. of moles of H2O formed in the reaction = 0.1 mole

Molecular weight of NaCl = 58.4 g/mol

Molecular weight of H20 = 18 g/mol

Therefore, gram of of NaCl formed in the reaction = 0.1*58.4 = 5.84 gram

Therefore, gram of of H2O formed in the reaction = 0.1*18 = 1.8 gram

Initial Water available = 50 + 50 = 100 ml = 100 g (Because Water density =1g/ml)

Therefore, Total water in the final mixture = 100+1.8 = 101.8 g = 101.8 ml

Therefore, Total final mass = Mass of NaCl + Mass of Water = 101.8+5.84 = 107.64 gm

Given, density of final mass = 1.04 g/ml

Therefore, Total final volume = 107.64/1.04 = 103.5 ml

Therefore, Molarity of NaCl = Moles of NaCl/Total Volume = 0.1/103.8/1000 = 0.97=~1 Molar

Given, Cp of 1 M NaCl solution = 3.93 J/g/K

Initial Temperature, Ti = 16.9 degC

Final Temperature, To = 30.4 degC

Therefore, heat absorbed by NaCl solution = m*Cp*(To-Ti) = 107.64*3.93*(30.4-16.9) = 5710 Joule = 5.71 KJ

Given, Total heat released by reaction = 57.32 KJ

Therefore, Heat absorbed by Calorimeter = (57.32-5.71) =51.61 KJ

Heat absorbed by Calorimeter = Calorimeter Heat capacity * (To-Ti)

Calorimeter Heat capacity = 51.61/(30.4-16.9) = 3.82 KJ/K

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