Question

PLZ SEE ATTACHED AND I WOULD REALLY APPRECIATE IT! ANYONE GOOD WITH CHEM

Answer 1

Answer : The value of $\Delta H_{rxn}$ for the reaction is, -390.3 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 main chemical reaction is:

$CH_4(g)+4Cl_2(g)\rightarrow CCl_4(g)+4HCl(g)$    $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C(s)+2H_2(g)\rightarrow CH_4(g)$     $\Delta H_1=-74.6kJ$

(2) $C(s)+2Cl_2(g)\rightarrow CCl_4(g)$     $\Delta H_2=-95.7kJ$

(3) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$     $\Delta H_2=-184.6kJ$

Now we are reversing reaction 1, multiplying reaction 3 by 2 and then adding all the equations, we get :

(1) $CH_4(g)\rightarrow C(s)+2H_2(g)$     $\Delta H_1=74.6kJ$

(2) $C(s)+2Cl_2(g)\rightarrow CCl_4(g)$     $\Delta H_2=-95.7kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$     $\Delta H_2=2\times (-184.6kJ)=-369.2kJ$

The expression for enthalpy change for the reaction will be,

$\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(74.6kJ)+(-95.7kJ)+(-369.2kJ)$

$\Delta H_{rxn}=-390.3kJ$

Therefore, the value of $\Delta H_{rxn}$ for the reaction is, -390.3 kJ