The formation of I2O5 & formation of 5CO + the reaction of I2O5 + 5CO = The formation of 5CO2 from hess cycle
Formation of I2O5 (-158) + the formation of 5CO (-550) is -708. Formation of 5CO2 is -1970 To find enthalpy of reaction it's (formation of 5CO2) - (formation of I2O5 + formation of 5CO) if you rearrange -1970 - (-708) = -1262 kJmol-1
Bonds in reactants: There is one N=N (I cant type triple bond just imagine it is one) 3 x H-H bonds
so enthalpy for reactants is (3 x 436) + 941 = 2249
Reactants - products = enthalpy change enthalpy for prods is 6 x N-H bonds (3 N-H bonds in each ammonia, and 2 molecules makes 6 bonds) Need to find N-H bond enthalpy
The formation of I2O5 & formation of 5CO + the reaction of I2O5 + 5CO = The formation of 5CO2 from hess cycle
Formation of I2O5 (-158) + the formation of 5CO (-550) is -708. Formation of 5CO2 is -1970 To find enthalpy of reaction it's (formation of 5CO2) - (formation of I2O5 + formation of 5CO) if you rearrange -1970 - (-708) = -1262 kJmol-1
Thank you so much! I was just confused as they used delta r h which means enthalpy change of reaction? No clue why they put that there then 😭 but thanks!
Bonds in reactants: There is one N=N (I cant type triple bond just imagine it is one) 3 x H-H bonds
so enthalpy for reactants is (3 x 436) + 941 = 2249
Reactants - products = enthalpy change enthalpy for prods is 6 x N-H bonds (3 N-H bonds in each ammonia, and 2 molecules makes 6 bonds) Need to find N-H bond enthalpy