Question

N2(8) + 3F2(g) → 2 NF3() AH 298 = - 264 kJ mol-'; AS98 = - 278 J K-'mol-' The following questions relate to the synthesis reaction represented by the chemical equation in the box above. a. Calculate the value of the standard free energy change, AG 298. for the reaction. b. Determine the temperature at which the equilibrium constant, Keg, for the reaction is equal to 1.00. (Assume that Hº and AS are independent of temperature.) c. Calculate the standard enthalpy change, AH', that occurs when a 0.256 mol sample of NF3() is formed from N2@) and F2) at 1.00 atm and 298 K The enthalpy change in a chemical reaction is the difference between energy absorbed in breaking bonds in the reactants and energy released by bond formation in the products. d. How many bonds are formed when two molecules of NF3 are produced according to the equation in the box above? e. Use both the information in the box above and the table of average bond enthalpies below to calculate the average enthalpy of the F- Fbond. Bond N N N-F F-F Average Bond Enthalpy (kJ mol!) 946 272 ?

Answer 1

a. ΔG0298 = ΔH0298 - T* ΔS0298

   ΔH0298 = -264 kJ / mole , T = 298 K and ΔS0298 = -278 J / K mole = -0.278 kJ / K mole

ΔG0298 = -264 - 298*-0.278 = -181.156 kJ / mole

b.  ΔG0rxn = -RT lnKeq

Here, Keq = 1 ⇒ lnKeq = ln 1 = 0 ⇒ -RT lnKeq = 0 ⇒   ΔG0rxn = -RT lnKeq = 0

So,  ΔG0rxn = ΔH0 - T* ΔS0 = 0

-264 - T*-0.278 = 0 ⇒ T = 264/0.278 = 949.64 K

c. Standard enthalpy change for the formation of 2 moles of NF3 = -264 kJ / mole

So, Standard enthalpy change for the formation of 1 moles of NF3 = -264/2 kJ / mole = -132 kJ / mole

So, Standard enthalpy change for the formation of 0.256 moles of NF3 = -132*0.256 = - 33.792 kJ  

d.Initially there are N2 and F2 only. So, one molecule of NF3 will form 3 new N-F bonds. So, 2 molecule of NF3 will form 2*3 = 6 new N-F bonds

e. ΔH0rxn = E(bonds breaking ) - E(bond-forming)

-264 kJ/ mole  = E (N≡N) + 3* E(F-F) - 2*3* E(N-F)

-264 kJ / mole = 946 kJ / mole + 3* E(F-F) - 6*272 kJ / mole

So, 3* E(F-F) = -264 +6*272 -946 = 422 kJ  

  E(F-F) = 422/3 = 140.66 kJ / mole