Write a balanced equation for the formation of copper(II) oxide at standard state. Include the enthalpy change as an energy term in the balanced equation.
Write a balanced equation for the simple decomposition of iron(II) oxide at standard state. Include the enthalpy change as an energy term in the balanced equation.
5 years ago
Answered By Leonardo F
For the first reaction, copper(II) oxide's formula is CuO, because oxygen has a -2 charge. The formation of a chemical compound is the reaction of production of this compound from its elements (how they occur in nature). The enthalpy of formation of copper and oxygen alone is zero, because they are simple substances, with only one chemical element. The standard enthalpy of formation for copper(II) oxide is approximately -155.2 kJ/mol. Since in the reaction we will have the production of 2 moles, we will have the release of 310.4 kJ of energy.
$2Cu+O_2\rightarrow2CuO$2Cu+O2→2CuO + 310.4 kJ
For the decomposition of iron(II) oxide, its formula is FeO. FeO is thermodynamically unstable below 575°C, tending to disproportionate to metal and $Fe_3O_4$Fe3O4 . The standard molar enthalpy of formation of FeO is -264.8 kJ/mol. The chemical reaction is:
$4FeO\rightarrow Fe+Fe_3O_4$4FeO→Fe+Fe3O4
The standard molar enthalpy of formation of $Fe_3O_4$Fe3O4 is -1120.9 kJ/mol. Calculating the enthalpy variation, we will have:
5 years ago
Answered By Leonardo F
For the first reaction, copper(II) oxide's formula is CuO, because oxygen has a -2 charge. The formation of a chemical compound is the reaction of production of this compound from its elements (how they occur in nature). The enthalpy of formation of copper and oxygen alone is zero, because they are simple substances, with only one chemical element. The standard enthalpy of formation for copper(II) oxide is approximately -155.2 kJ/mol. Since in the reaction we will have the production of 2 moles, we will have the release of 310.4 kJ of energy.
$2Cu+O_2\rightarrow2CuO$2Cu+O2→2CuO + 310.4 kJ
For the decomposition of iron(II) oxide, its formula is FeO. FeO is thermodynamically unstable below 575°C, tending to disproportionate to metal and $Fe_3O_4$Fe3O4 . The standard molar enthalpy of formation of FeO is -264.8 kJ/mol. The chemical reaction is:
$4FeO\rightarrow Fe+Fe_3O_4$4FeO→Fe+Fe3O4
The standard molar enthalpy of formation of $Fe_3O_4$Fe3O4 is -1120.9 kJ/mol. Calculating the enthalpy variation, we will have:
Enthalpy variation = (-264.8)(-4) + (1)(-1120.9)
Enthalpy variation = -61.7 kJ (releases energy)
Hence:
$4FeO\rightarrow Fe+Fe_3O_4+61.7kJ$4FeO→Fe+Fe3O4+61.7kJ