Consider the following intermediate chemical equations.
CH2(g) →C(s) + 2H2(g)
CC1.(g) → C(s) +2012(g)
H2(g) +C12(g) → 2HCl(9)
ΔH, = 74.6 kJ
ΔH2 = 95.7 kJ
ΔΗ =
ΔHg =-92.3 kJ
What is the enthalpy of the overall chemical reaction CH (g) +4012(g) → CCl4(h)+ 4HCl(g)?

Respuesta :

Answer: The enthalpy of the overall chemical equation is -205.7 kJ

Explanation:

Hess's law is defined as the law which states that the total amount of heat absorbed or released for a given chemical reaction will remain the same whether the process occurs in one step or several steps.

In this law, a chemical equation is treated as an ordinary algebraic expression that can be added or subtracted to yield the required overall equation.  

The enthalpy change or heat of the overall reaction is the summation of the enthalpies or heats of the intermediate reactions involved in the process.

The given overall chemical equation follows:

[tex]CH_4(g)+4Cl_2(g)\rightarrow CCl_4(g)+4HCl(g);\Delta H^o_{rxn}[/tex]

The intermediate equations for the above reaction are:

(1)  [tex]CH_4(g)\rightarrow C(s)+2H_2(g);\Delta H_1=74.6 kJ[/tex]

(2)  [tex]CCl_4(g)\rightarrow C(s)+2Cl_2(g);\Delta H_2=95.7kJ[/tex]

(3)  [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g);\Delta H_3=-92.3kJ[/tex]

According to Hess law, the enthalpy of the reaction becomes:

[tex]\Delta H^o_{rxn}=[1\times (\Delta H_1)] + [1\times (-\Delta H_2)] + [2\times (\Delta H_3)][/tex]

Putting values in above expression, we get:

[tex]\Delta H^o_{rxn}=[1\times (74.6)] + [1\times (-95.7)] + [2\times (-92.3)]\\\\\Delta H^o_{rxn}=-205.7kJ[/tex]

Hence, the enthalpy of the overall chemical equation is -205.7 kJ