Question

Question 3 (1 point) Given the following reactions and their corresponding changes in enthalpy 2Al(s) + 3Br (1) Z 2A1Br,(s) Ahrº=-1054 kJmol'i SnBr(s) 2 Sn(s) + 2Br (1) Ahrº=377.4 kJmoli What is the change in enthalpy for the following reaction in kJmoll? 2Al(s) + Br_(1) + SnBr,(s) 32A1Br (s) + Sn(s) Ahrº = O 1) -1809 2) -677 3) -1431 4) 2485 5) 3917

I hope the answer is clear and correct thanks

Answer 1

a 5 path 2 same. F g According to How low, the enthalpy change accompanying particular reaction is independent of path followed and depends Only on initial and jinal states. for example for transition A to B opathe =D> for all the paths A enthalpy g seection is path3 -56-H Applications Hess Law in 1) Chemical reactions can be added or subtracted along with their enthalpy values. 2) the mical reactions can be multiplied or divided by some constant along with the same operation on the enthalpy values, 3) A Chemical reaction can be reversed the sign o, enthalpy values.? suevering now look at given probleme 3) 2 Al(s) + 3Br2(e) 2A1Br 3(5) įsti = -1054 kJ mott Sn(s) + 2Br2(e) Drir = 377.4 kJ molt A close inspection reveals that the given reaction can be obtained by adding above 2 reactions: 2 A1(8) +3 Brace) +Sn Bryl) = 2A1BY, (s) +Sn(s)+2Br(e); -18=4-1054+3774) kJmoly 2Al(s) + Br, (1) + Sn Bry() = 2 AlBrz (s) + Sn(s); H-676.6 kJ mol- along with Let us Sn Bry (5)

So, the correct option is - 677 kJ molt 4) 2Al(s) +3Bryles ² 2 Albr, (S) ; Dhr = 1054 kJ molt 2A1FC 2Al(s) + 3529) ; DH = 300 8.2 kJ molt Again, adding above two reactions : 2 Arts) +387,6) +2AJĘ(S) 201B72() +2016+3F,9); s4*= (-1054 +3068-2) 3 Br,le) + 2A) (5) 2 A1 Brzes) +37749) Ahir = 1954.2kJmol? So, the correct option is /1954 | kJ molt KJ molt you need to reverse poso you can do similar problems as shown above. Sometimes, a chemical reaction to bring the reactants and products to appropriate sides But for that , the sign of shoh also needs to be reversed. please rate