Respuesta :
Answer : The amount of heat involved in the complete reaction is -367.5 kJ
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]Pb(s)+C(s)+\frac{3}{2}O_2\rightarrow PbCO_3(s)[/tex] [tex]\Delta H=?[/tex]
The intermediate balanced chemical reaction will be,
(1) [tex]Pb(s)+\frac{1}{2}O_2(g)\rightarrow PbO(s)[/tex] [tex]\Delta H_1=-219kJ[/tex]
(2) [tex]C(g)+O_2(g)\rightarrow CO_2(g)[/tex] [tex]\Delta H_2=-394kJ[/tex]
(3) [tex]PbCO_3(s)\rightarrow PbO(s)+CO_2(g)[/tex] [tex]\Delta H_3=86kJ[/tex]
Now we will reverse the reaction 3 and then adding all the equations, we get :
(1) [tex]Pb(s)+\frac{1}{2}O_2(g)\rightarrow PbO(s)[/tex] [tex]\Delta H_1=-219kJ[/tex]
(2) [tex]C(g)+O_2(g)\rightarrow CO_2(g)[/tex] [tex]\Delta H_2=-394kJ[/tex]
(3) [tex]PbO(s)+CO_2(g)\rightarrow PbCO_3(s)[/tex] [tex]\Delta H_3=-86kJ[/tex]
The expression for enthalpy of change will be,
[tex]\Delta H=\Delta H_1+\Delta H_2+\Delta H_3[/tex]
[tex]\Delta H=(-219kJ)+(-394kJ)+(-86kJ)[/tex]
[tex]\Delta H=-699kJ[/tex]
Now we have to calculate the moles of carbon.
[tex]\text{Moles of carbon}=\frac{\text{Mass of carbon}}{\text{Molar mass of carbon}}[/tex]
Molar mass of carbon = 12 g/mole
[tex]\text{Moles of carbon}=\frac{6.309g}{12g/mole}=0.5258mole[/tex]
Now we have to calculate the amount of heat involved in the complete reaction.
As, 1 mole of carbon involved heat = -699 kJ
So, 0.5258 moles carbon involved heat = 0.5258 × (-699 kJ) = -367.5 kJ
Therefore, the amount of heat involved in the complete reaction is -367.5 kJ
Based on the enthalpy values provided, the amount of heat involved in the complete reaction is -367.5 kJ
What is enthalpy change of a reaction?
The enthalpy change of a reaction is the amount of heat evolved or absorbed when reactant molecules react to form products.
The given reaction occurs in several intermediate steps.
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given multistep chemical reaction is constant and is independent of the reaction pathway.
The given main reaction is,
Pb (s) + C (s) + 3/2 O2 (g) → PbCO3 ; ΔH = ?
The intermediate steps leading to the main reaction is as follows:
- Pb (s) + 1/2 O2 (g) → PbO ; ΔH1 = -219 kJ
- C (s) + O2 (g) → CO2 ; ΔH2 = -394 kJ
- PbCO3 → PbO (s) + CO2 (g) ; ΔH3 = 86 kJ
- Reversing reaction 3;
- PbO (s) + CO2 (g) → PbCO3. ; ΔH4 = -86 kJ
Combining equations 1, 2 and 4 gives the main equation
- Pb (s) + C (s) + 3/2 O2 (g) → PbCO3 ; ΔH = ?
Enthalpy change ΔH = ΔH1 + ΔH2 + ΔH4
ΔH = (-219 kJ) + (-394 kJ) + (-86 kJ)
ΔH = -699 kJ / mol
Moles of carbon in 6.309 g = mass/molar mass
Molar mass of carbon = 12 g/mole
Moles of carbon = 6.309g/12g/mol
Moles of carbon = 0.5258mole
Thus, amount of heat involved in the complete reaction = 0.5258 × (-699 kJ) = -367.5 kJ
Therefore, the amount of heat involved in the complete reaction is -367.5 kJ
Learn more about enthalpy and Hess law at: https://brainly.com/question/9328637
