Respuesta :
Answer:
The standard enthalpy of formation of 1 mol of calcium carbonate is -1207.6 kJ/mol.
Explanation:
[tex]Ca(OH)_2(s) \rightarrow CaO(s) + H_2O(l) \Delta H^o = 65.2 kJ/mol[/tex]..[1]
[tex]Ca(OH)_2(s) + CO_2(g) \rightarrow CaCO_3(s) + H_2O(l) \Delta H^o = -113.8 kJ/mol [/tex]...[2]
[tex]C(s) + O_2(g) \rightarrow CO_2(g). \Delta H^o = -393.5 kJ/mol [/tex]..[3]
[tex]2 Ca(s) + O_2(g) \rightarrow 2 CaO(s). \Delta H^o = -1270.2 kJ/mol [/tex]..[4]
[tex]2Ca(s)+2C(s)+3O_2(g)\rightarrow 2CaCO_3(s),\Delta H^{o}_f = ?[/tex]...[5]
2× [2] + 2 × [3] + [4] - 2 × [1] = [5] (Hess's law)
[tex]\Delta H^{o}_f=2\times (-113.8 kJ/mol)+2\times (-393.5 kJ/mol)+(-1270.2 kJ/mol)-2\times (65.2 kJ/mol)[/tex]
[tex]\Delta H^{o}_f=-2,415.2 kJ/mol[/tex]
The standard enthalpy of formation of 2 mol of calcium carbonate is -2,415.2 kJ/mol.
The standard enthalpy of formation of 1 mol of calcium carbonate :
[tex]\frac{-2,415.2 kJ/mol}{2}= -1207.6 kJ/mol[/tex]
The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals
The correct answer is -1207.6kj/mole.
What is enthalpy?
- Enthalpy a property of a thermodynamic system, is the sum of the system's internal energy and the product of its pressure and volume.
- It is a state function used in many measurements in chemical, biological, and physical systems at a constant pressure, which is conveniently provided by the large ambient atmosphere.
According to the question, the Hf will be:-
[tex]2*(-113.8)+2*(393.5)+(-1270.2)-2(65.2)[/tex]
The hf will be -2415.2kj
The standard enthalpy of formation of 2 moles of calcium carbonate is -2,415.2 kJ/mol.
The standard enthalpy of formation of 1 mole of calcium carbonate:
[tex]\frac{-2415}{2} \\= -1207.9kj[/tex]
Hence, the correct answer to -1207.9kj.
For more information about the hf, refer to the link:-
https://brainly.com/question/7932885
