Respuesta :
Answer: The [tex]\Delta H^o_{rxn}[/tex] for the reaction is 72 kJ.
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:
[tex]2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)[/tex] [tex]\Delta H^o_{rxn}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]2Sr(s)+O_2(g)\rightarrow 2SrO(s)[/tex] [tex]\Delta H_1=-1184kJ[/tex]
(2) [tex]SrO(s)+CO_2(g)\rightarrow SrCO_3(s)[/tex] [tex]\Delta H_2=-234kJ[/tex] ( × 2)
(3) [tex]CO_2(g)\rightarrow C(s)+O_2(g)[/tex] [tex]\Delta H_3=394kJ[/tex] ( × 2)
The expression for enthalpy of the reaction follows:
[tex]\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ[/tex]
Hence, the [tex]\Delta H^o_{rxn}[/tex] for the reaction is 72 kJ.
Considering the Hess's Law, the enthalpy change for the reaction is 2440 kJ.
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:
2 SrCO₃ (s) → 2 Sr (s) + 2 C(s) + 3 O₂ (g)
which occurs in three stages.
You know the following reactions, with their corresponding enthalpies:
Equation 1: 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol
Equation 2: SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol
Equation 3: CO₂ (g) → C(s) + O₂ (g) ∆H° = 394 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 2 moles of SrCO₃ (s) on reactant side and it is present in second equation.
Since this equation has 1 mole of SrCO₃ (s) on the product side, it is necessary to locate the SrCO₃ (s) on the reactant side (invert it) and multiply it by 2 to obtain 2 moles of SrCO₃ (s) on the reactant side.
When an equation is inverted, the sign of ΔH also changes. And since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiplied by 2, the variation of enthalpy also.
- Second step
Now, Sr (s) must be a product and is present in the first equation. Since this equation has 2 moles of Sr (s) on the reactant side, it is necessary to locate the Sr (s) on the reactant side (invert it) ans the ΔH also changes.
- Third step
Finally, 2 moles of C (s) must be a product and is present in the third equation. Since this equation has 1 mole of C (s) on the product side, it is necessary multiply it by 2 to obtain 2 moles of C (s). Since enthalpy is an extensive property, since the equation is multiplied by 2, the variation of enthalpy also.
In summary, you know that three equations with their corresponding enthalpies are:
Equation 1: 2 SrO (s) → 2 Sr(s) + O₂ (g) ∆H° = 1184 kJ/mol
Equation 2: 2 SrCO₃ (s) → 2 SrO (s) + 2 CO₂ (g) ∆H° = 468 kJ/mol
Equation 3: 2 CO₂ (g) → 2 C(s) + 2 O₂ (g) ∆H° = 788 kJ/mol
Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
2 SrCO₃ (s) → 2 Sr (s) + 2 C(s) + 3 O₂ (g) ΔH= 2440 kJ
Finally, the enthalpy change for the reaction is 2440 kJ.
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