How to Calculate Enthalpy Change for a Zinc and Copper Sulfate Reaction?

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SUMMARY

The discussion focuses on calculating the enthalpy change for the reaction between zinc (Zn) and copper(II) sulfate (CuSO4). The experiment involved adding 5.96 grams of zinc to 25 cm³ of 1M copper(II) sulfate, resulting in a temperature drop from 68°C to 19.5°C. The calculated enthalpy change (ΔH) for the reaction is -55880 J/mol, indicating an exothermic reaction. The thermochemical equation derived from the calculations is Zn + Cu²⁺ → Zn²⁺ + Cu, with ΔH = -55880 J/mol.

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Homework Statement



Zn(solid) + Cu2+(aq) -> Cu(solid) + Zn2+(aqueous)
5.96grams on zinc was added to 25cm^3 of 1M copper(II)sulphate solution in a polystyrene cup.
Over 6 and a half minutes, the temperature of the solution decreased from 68 degrees Celsius to 19.5 degrees C.

1. Calculate enthalpy change for the quantities used, making appropriate assumptions
2. Calculate the enthalpy change for one mole of Zn and CuSO4(aq) and write the thermochemical equation for the reaction

2. The attempt at a solution

1.
moles of Zn = 5.96/65.4 = 0.09113... or 149/1635
therefore, moles of Cu2+ in CuSO4 = 149/1635 (because ratio in above equation is 1 to 1


mass of CuSO4 in solution = 149/1635 x molar mass of CuSO4 = 14.54 grams

q = mcDT,
m = 25 grams, c = 4.2, DT = 68-19.5 = 48.5
q = 5092.5 joules

DH = q/n
= 5092.5/(149/1635)
= 55880Jmol^-1

temperature increased so enthalpy is negative
-> -55880joules per mol

2. Zn + Cu2+ -> Zn2+ + Cu DH = -55880 joules



I'm not sure if I'm doing this right...
 
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mgnymph said:
2. The attempt at a solution

1.
moles of Zn = 5.96/65.4 = 0.09113... or 149/1635
therefore, moles of Cu2+ in CuSO4 = 149/1635 (because ratio in above equation is 1 to 1

For the moles of Cu2+, you'll need to use the information given about the copper(II)sulphate solution. It's not necessarily the same # of moles as the Zinc ... either Cu or Zn will be the limiting reactant here.

Other than that, your methods look fine.
 

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