Respuesta :
Answer : The enthalpy of the given reaction will be, -1048.6 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.
The main reaction is:
[tex]XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)[/tex] [tex]\Delta H=?[/tex]
The intermediate balanced chemical reactions are:
(1) [tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)[/tex] [tex]\Delta H_1=-241.8kJ[/tex]
(2) [tex]X(s)+2Cl_2(g)\rightarrow XCl_4(s)[/tex] [tex]\Delta H_2=+461.9kJ[/tex]
(3) [tex]\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)[/tex] [tex]\Delta H_3=-92.3kJ[/tex]
(4) [tex]X(s)+O_2(g)\rightarrow XO_2(s)[/tex] [tex]\Delta H_4=-789.1kJ[/tex]
(5) [tex]H_2O(g)\rightarrow H_2O(l)[/tex] [tex]\Delta H_5=-44.0kJ[/tex]
Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :
(1) [tex]2H_2O(g)\rightarrow 2H_2(g)+O_2(g)[/tex] [tex]\Delta H_1=2\times 241.8kJ=483.6kJ[/tex]
(2) [tex]XCl_4(s)\rightarrow X(s)+2Cl_2(g)[/tex] [tex]\Delta H_2=-461.9kJ[/tex]
(3) [tex]2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)[/tex] [tex]\Delta H_3=4\times -92.3kJ=-369.2kJ[/tex]
(4) [tex]X(s)+O_2(g)\rightarrow XO_2(s)[/tex] [tex]\Delta H_4=-789.1kJ[/tex]
(5) [tex]2H_2O(l)\rightarrow 2H_2O(g)[/tex] [tex]\Delta H_5=2\times 44.0kJ=88.0kJ[/tex]
The expression for enthalpy of main reaction will be:
[tex]\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5[/tex]
[tex]\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)[/tex]
[tex]\Delta H=-1048.6kJ[/tex]
Therefore, the enthalpy of the given reaction will be, -1048.6 kJ
