Question

1) When 0.119 g of Zn(s) is combined with enough HCl to make 51.6 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.4 ∘C to 23.9 ∘C. Find ΔHrxn for this reaction as written. (Use 1.0 g/mL for the density of the solution and 4.18 J/g⋅∘C as the specific heat capacity.)

2)Calculate ΔHrxn for the following reaction:

CaO(s)+CO2(g)→CaCO3(s)

Use the following reactions and given ΔH values:

Ca(s)+CO2(g)+12O2(g)→CaCO3(s), ΔH= -812.8 kJ
2Ca(s)+O2(g)→2CaO(s), ΔH= -1269.8 kJ

Answer 1

1)

total volume of solution = 51.6 ml

solution density = 1 g/mm

solution mass = density x volume = 1 x 51.6 = 51.6g

q = m Cp dT

    = 51.6 x 4.184 x (23.9-22.4)

     = 323.84 J

moles of Zn = 0.119 / 65.38 = 1.82 x 10^-3

1 mol     ----------------> 323.84 J

1.82 x 10^-3 mol -------------> 323.84 x 1.82 x 10^-3 = 0.589 J

DHrxn = 5.89 x 10^-4 kJ/mol

2)

Calculate ΔHrxn for the following reaction:

CaO(s)+CO2(g)→CaCO3(s)

Ca(s) + CO2(g) + 1/2 O2(g)------------------>CaCO3(s),     ΔH= -812.8 kJ -----------------> 1

2Ca(s)+O2(g)-------------------> 2CaO(s),   ΔH= -1269.8 kJ ------------------------> 2

do the reverse of second eqation and divide with 2 and add to 1st eqaution

CaO (s) ------------------> Ca(s)   + 1/2 O2 (g) , ΔH= 634.9 kJ ------------------------> 3

add (1) + (3)

Ca(s) + CO2(g) + 1/2 O2(g)------------------>CaCO3(s),     ΔH= -812.8 kJ -----------------> 1

CaO (s) ------------------> Ca(s)   + 1/2 O2 (g) ,                     ΔH= 634.9 kJ ------------------------> 3

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CaO(s)+CO2(g)----------------------------> CaCO3(s) , ΔH = -177.9 kJ