How do you find the #ΔH# of the following reaction: #SnCl_2(s) + Cl_2(g) → SnCl_4(l)#?

Given the following data:
#SnCl_2(s) + TiBr_2(s) → SnBr_2(s) + TiCl_2(s)#
#ΔH = +4.2 kJ#
#TiBr_2(s) + SnCl_4(l) → SnBr_2(s) + TiCl_4(l)#
#ΔH = -74 kJ#
#TiCl_4(l) → TiCl_2(s) + Cl_2(g)#
#ΔH = +273 kJ#

1 Answer
Apr 23, 2016

Answer:

#DeltaH=-194.8kJ#

Explanation:

Knowing that elemental molecules (made of only one element, like #O_2, Cl_2, Br_2#) etc have #0# enthalpy, and the equation for enthalpy

#DeltaH_("rxn")=Sigman*"products"-Sigman*"reactants"#,

or overall enthalpy (#DeltaH_("rxn")#) is the enthalpy of products minus enthalpy of reactants, we can rearrange equation #3.#.

#TiCl_4->TiCl_2+Cl_2, DeltaH=+273kJ#

#273=(TiCl_2+Cl_2)-TiCl_4# (products minus reactants)

#273=TiCl_2-TiCl_4# (enthalpy of #Cl_2# is #0#)

#TiCl_4=TiCl_2-273# (rearranged algebraically)

Now you can substitute this into equation #2.#, again using the products minus reactants equation

#TiBr_2+SnCl_4->SnBr_2+TiCl_4, DeltaH=-74kJ#

#-74=(SnBr_2+TiCl_4)-(TiBr_2+SnCl_4)#

#-74=SnBr_2+TiCl_2-273-TiBr_2-SnCl_4# (substituting in equation #3.# and expanding out everything)

#199=SnBr_2+TiCl_2-TiBr_2-SnCl_4# (rearranging, adding #273# to both sides)

You also have equation #1.#, which you can use products minus reactants to find

#SnCl_2+TiBr_2->SnBr_2+TiCl_2, DeltaH=+4.2kJ#

#4.2=(SnBr_2+TiCl_2)-(SnCl_2+TiBr_2)# (products minus reactants)

#4.2=SnBr_2+TiCl_2-SnCl_2-TiBr_2# (expanding out brackets)

Now you can equate equations #1.# and #2.# in their new form, as so

#SnBr_2+TiCl_2-SnCl_2-TiBr_2+194.8=SnBr_2+TiCl_2-TiBr_2-SnCl_4#

because if you add #194.8# to equation #1# it equals #199# in total, the same as equation #2.# With these two equal to each other, you can cancel out the similarities on both sides and find what's left.

Both sides have a #SnBr_2#, so you can cancel that. They also have a #TiCl_2# and a #-TiBr_2#, so get rid of those from each side, leaving

#-SnCl_2+194.8=-SnCl_4#

#SnCl_4-SnCl_2=-194.8kJ#

This is the answer to the question in the form of products minus reactants, because you can see the product is #SnCl_4# and the reactants are #SnCl_2# and #Cl_2#, though #Cl_2# has an enthalpy of #0# so it can be ignored.