The molar entropies of HI(g), H(g) and I(g) at 298 K are 206.5, 114.6, and 180.7 J mol^(-1)K^(-1)respectively. Using the DeltaG^(@) given below, calculate the bond energy of HI.HI(g)rarrH(g)+I(g)," "DeltaG^(@)=271.8 kJ

The molar entropies of HI(g), H(g) and I(g) at 298 K are 206.5, 114.6,

1.	The molar entropies of HI(g), H(g) and I(g) at 298 K are 206.5, 114.6, and 180.7 J mol^(-1)K^(-1)respectively. Using the DeltaG^(@) given below, calculate the bond energy of HI.HI(g)rarrH(g)+I(g)," "DeltaG^(@)=271.8 kJ
Answer» `282.4 kJ "mol"^(-1)` `298.3 kJ "mol"^(-1)` `290.1 kJ "mol"^(-1)` `315.4 kJ "mol"^(-1)` Solution :`DeltaS^(@) =-206.5+114.6 +180.7 =88.8` `DeltaG^(@)=DELTAH^(@)-TDeltaS^(@)` `DeltaH^(@)=271.8 + 298 xx88.8 XX 10^(-3)` `DeltaH^(@)=298.3`