What is the enthalpy change to roast one mol of copper 2 sulfide?
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.
To determine the enthalpy change to roast one mole of copper(II) sulfide (CuS), we need to consider the balanced chemical equation for the reaction.
The balanced chemical equation for the roasting of copper(II) sulfide is:
2 CuS(s) + 3 O2(g) → 2 CuO(s) + 2 SO2(g)
The coefficients in this equation represent the mole ratios between the reactants and products. From the reaction equation, we can see that the stoichiometric coefficient of CuS is 2.
To calculate the enthalpy change (ΔH) for roasting one mole of CuS, we need to know the enthalpy change of the reaction per mole of CuS.
Unfortunately, the enthalpy change for this specific reaction is not readily available. To obtain the enthalpy change, experimental data or thermodynamic calculations would be needed. Without such data, the enthalpy change cannot be determined accurately.
However, if the enthalpy change data were available, it would be given in units of either kilojoules (kJ) or joules (J) per mole of CuS.
To determine the enthalpy change to roast one mole of copper(II) sulfide (CuS), we need to refer to the relevant thermochemical equation and corresponding enthalpy values.
1. The thermochemical equation for the roasting of copper(II) sulfide is:
CuS(s) + O2(g) → CuO(s) + SO2(g)
2. To calculate the enthalpy change (∆H) for this reaction, you need the enthalpy of formation values (ΔHf) for each compound involved in the reaction. Here are the values:
ΔHf(CuS) = -52.5 kJ/mol
ΔHf(CuO) = -156.2 kJ/mol
ΔHf(SO2) = -296.8 kJ/mol
3. With these values, you can calculate the enthalpy change using the following equation:
∆H = ∑ΔHf(products) - ∑ΔHf(reactants)
∆H = [ΔHf(CuO) + ΔHf(SO2)] - [ΔHf(CuS) + ΔHf(O2)]
Substitute the values into the equation:
∆H = [-156.2 kJ/mol + (-296.8 kJ/mol)] - [-52.5 kJ/mol + 0 kJ/mol]
∆H = -452.9 kJ/mol + 52.5 kJ/mol
∆H = -400.4 kJ/mol
Therefore, the enthalpy change to roast one mole of copper(II) sulfide is -400.4 kJ/mol. The negative sign indicates that the reaction is exothermic, releasing energy.