HELP ASAP: Determine the enthalpy change of the following reaction: CH4 + 4Cl2 -> CCl4+ 4HCl  Given enthalpies: 
CH4: -17.9 kJ/mol 
Cl2: 0 kJ/mol 
CCl4: -95.98 kJ/mol 
HCl: -92.3 kJ/mol

A. -447.28 kJ/mol

B. -206.18

C. -170.38

D. -393.58

A

Explanation:

In this question, we are to calculate the enthalpy of change of the reaction. ΔH

To be able to do that, we use the Hess’ law and it involves the subtraction of the summed heat reaction of the reactants from that of the product.

Thus, mathematically, the enthalpy of change of the reaction would be;

[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4) + 4 ΔH(Cl2)]

We can see that we multiplied some heat change by some numbers. This is corresponding to the number of moles of that compound in question in the reaction.

Also, for diatomic gases such as chlorine in the reaction above, the heat of reaction is zero.

Thus, we can have the modified equation as follows;

[ΔH(CCl4) + 4 ΔH(HCl)] - [ΔH(CH4)]

Substituting the values we have according to the question, we have;

-95.98 + 4(-92.3) -(-17.9)

= -95.98 - 369.2 + 17.9

= -447.28 KJ/mol

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

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