Respuesta :
Answer : The enthalpy change for the decomposition of calcium carbonate is, 178.1 kJ/mol
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given main reaction is,
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex] [tex]\Delta H=?[/tex]
The intermediate balanced chemical reaction will be,
(1) [tex]Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)[/tex] [tex]\Delta H_1=65.02kJ/mol[/tex]
(2) [tex]Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)[/tex] [tex]\Delta H_2=-113.8kJ/mol[/tex]
(3) [tex]C(s)+O_2(g)\rightarrow CO_2(g)[/tex] [tex]\Delta H_3=-393.5kJ/mol[/tex]
(4) [tex]2Ca(s)+O_2(g)\rightarrow 2CaO(s)[/tex] [tex]\Delta H_4=-1270.2kJ/mol[/tex]
Now we are reversing reaction 1 and then adding reaction 1 and 2, we get :
(1) [tex]Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)[/tex] [tex]\Delta H_1=65.02kJ/mol[/tex]
(2) [tex]CaCO_3(s)+H_2O(l)\rightarrow Ca(OH)_2(s)+CO_2(g)[/tex] [tex]\Delta H_2=113.8kJ/mol[/tex]
The expression for enthalpy of change will be,
[tex]\Delta H=\Delta H_1+\Delta H_2[/tex]
[tex]\Delta H=(65.02)+(113.08)[/tex]
[tex]\Delta H=178.1kJ/mol[/tex]
Thus, the enthalpy change for the decomposition of calcium carbonate is, 178.1 kJ/mol
Considering the Hess's Law, the correct answer is option c.: the enthalpy change for the reaction is 179 kJ/mol.
Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.
In this case you want to calculate the enthalpy change of:
CaCO₃ (s) → CaO (s) + CO₂ (g)
You know the following reactions, with their corresponding enthalpies:
Equation 1: Ca(OH)₂ (s) → CaO (s) + H₂O (l) ΔH = 65.2 kJ/mol
Equation 2: Ca(OH)₂ (s) + CO₂ (g) → CaCO₃ (s) + H₂O (l) ΔH = –113.8 kJ/mol
Equation 3: C (s) + O₂ (g) → CO₂ (g) ΔH = -393.5 kJ/mol
Equation 4: 2 Ca(s) + O₂ (g) → 2 CaO (s) ΔH= -1270.2 kJ/mol
Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.
- First step
To obtain the enthalpy of the desired chemical reaction you need one mole of CaCO₃ (s) on reactant side and it is present in second equation. Since this equation has 1 mole of CaCO₃ (s) on the product side, it is necessary to locate the O on the reactant side (invert it). When an equation is inverted, the sign of ΔH also changes.
- Second step
Now, 1 mole of CaO (s) must be a product and is present in the first equation. So let's write this as such.
In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: Ca(OH)₂ (s) → CaO (s) + H₂O (l) ΔH = 65.2 kJ/mol
Equation 2: CaCO₃ (s) + H₂O (l) → Ca(OH)₂ (s) + CO₂ (g) ΔH = 113.8 kJ/mol
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
CaCO₃ (s) → CaO (s) + CO₂ (g) ΔH= 65.2 kJ/mol + 113.8 kJ/mol= 179 kJ/mol
Finally, the correct answer is option c.: the enthalpy change for the reaction is 179 kJ/mol.
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