Question
What is the enthalpy change to roast one mol of copper 2 sulfide?
Answers
GPT 3.5
The balanced equation for the roasting of one mole of copper(II) sulfide (CuS) is:
2 CuS(s) + 3 O2(g) -> 2 CuO(s) + 2 SO2(g)
To determine the enthalpy change for this reaction, we need to know the enthalpy of formation (ΔHf) for each compound involved.
The values of ΔHf for CuS, CuO, and SO2 are:
ΔHf(CuS) = -79.5 kJ/mol
ΔHf(CuO) = -156.2 kJ/mol
ΔHf(SO2) = -296.8 kJ/mol
Using these values, we can calculate the enthalpy change for the reaction:
2 ΔHf(CuS) + 3 ΔHf(O2) - 2 ΔHf(CuO) - 2 ΔHf(SO2)
= 2 (-79.5 kJ/mol) + 3(0 kJ/mol) - 2 (-156.2 kJ/mol) - 2 (-296.8 kJ/mol)
= -159 kJ/mol + 312.4 kJ/mol + 593.6 kJ/mol
= 746 kJ/mol
Therefore, the enthalpy change to roast one mole of copper(II) sulfide is approximately 746 kJ/mol.
2 CuS(s) + 3 O2(g) -> 2 CuO(s) + 2 SO2(g)
To determine the enthalpy change for this reaction, we need to know the enthalpy of formation (ΔHf) for each compound involved.
The values of ΔHf for CuS, CuO, and SO2 are:
ΔHf(CuS) = -79.5 kJ/mol
ΔHf(CuO) = -156.2 kJ/mol
ΔHf(SO2) = -296.8 kJ/mol
Using these values, we can calculate the enthalpy change for the reaction:
2 ΔHf(CuS) + 3 ΔHf(O2) - 2 ΔHf(CuO) - 2 ΔHf(SO2)
= 2 (-79.5 kJ/mol) + 3(0 kJ/mol) - 2 (-156.2 kJ/mol) - 2 (-296.8 kJ/mol)
= -159 kJ/mol + 312.4 kJ/mol + 593.6 kJ/mol
= 746 kJ/mol
Therefore, the enthalpy change to roast one mole of copper(II) sulfide is approximately 746 kJ/mol.