
Answer:
the enthalpy change using Hess's Law in the conversion = -592 KJ
Explanation:
If you clearly observe all three reactions mentioned in the question , you will notice that the product of the former reactions becomes the reactant of the later , such that the final product , i.e. pure Si (s) can be obtained by addition of all these three reactions.
SiOâ‚‚(s) + 2C(s) ------ Si(impure solid) + 2CO(g) Â +690
Si(impure solid) + Clâ‚‚(g) ======> SiClâ‚„(g) Â -657
SiClâ‚„(g) + Mg(s) =====> MgClâ‚‚(s) + Si(s) Â Â -625
So the enthalpy change for the overall reaction for the formation of pure Si will be given as :
= (+690) + (-657) + (-625)
= -592 KJ