Question

When an excess of Zn is added to 125mL of 0.150M CuSO4(aq) in a constant-pressure calorimeter of negligible heat capacity, the temperature of the solution rises from 21.20 C to 28.97 C. Assuming the density and specific heat of the solution are the same as for pure water (1.00 g/mL and 4.184 J/g C), determine the molar enthalpy change of the following reaction. Ignore the specific heats of the metals

Zn(s) + CuSO4(aq) --> ZnSO4 (aq) + Cu(s)

Answer 1

-216.75kJ/mol

Explanation

Density of solution = 1.00g/ml

So,

Mass of Solution = 125ml × 1.00g/ml = 125g

Heat absorbed by solution is q

q = m × ∆T × C

m = mass of solution, 125g

∆T = 28.97℃ - 21.20℃ = 7.77℃

C = heat capacity of solution, 4.184J/g℃

q = 125g × 7.77℃ × 4.184J/g℃

= 4064J

∆H for the reaction = -q = -4064J

now write the reaction

Zn(s) + CuSO4(aq) --------> ZnSO4(aq) + Cu(s)

this reaction is 1:1 molar reaction

no of mole of CuSO4 = (0.150mol/1000ml)×125ml = 0.01875mol

molar enthalphy change, ∆Hrxn = -4064J/0.01875mol = -216.75kJ/mol