Question

Given the following two reactions: (1)Cu(s) + Cl2(g)CuCl2(s) H(1) = -220.1 kJ (2)Ni(s) + Cl2(g)NiCl2(s) H(2)...

Given the following two reactions: (1)Cu(s) + Cl2(g)CuCl2(s) H(1) = -220.1 kJ (2)Ni(s) + Cl2(g)NiCl2(s) H(2) = -305.3 kJ calculate the enthalpy change for the following reaction: (3) Cu(s) + NiCl2(s) CuCl2(s) + Ni(s) H(3)

kJ =

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Answer #1

The exercise in question can be summarized as follows::

So if we apply the Hess law and carry out a sum of reactions 1 and 2, as well as an exchange of products to reactants and reactants to products in reaction number 2 with their respective sign change in enthalpy you have to:

Then the reaction change enthalpy for the third equation is 85,5 KJ

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Given the following two reactions: (1)Cu(s) + Cl2(g)CuCl2(s) H(1) = -220.1 kJ (2)Ni(s) + Cl2(g)NiCl2(s) H(2)...
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