Given the equation N2 + 3F2 --> 2NF3 and the values of deltaH (-264 kJ/mol) and deltaS (-278 J/mol K), what is the standard free energy change for this reaction? B) Use the rxn and table (below) to calculate the average enthalpy of the F-F bond.

bond average bond enthalpy
triple N bond 946
n - f 272
F - F ?

1 Answer
Nov 7, 2016

Answer:

#sf((a))#

#sf(-181color(white)(x)"kJ/mol")#

#sf((b))#

#sf(141color(white)(x)"kJ")#

Explanation:

#sf((a))#

The relationship you need is:

#sf(DeltaG^(@)=DeltaH^(@)-TDeltaS^(@))#

#:.##sf(DeltaG^(@)=-264xx10^(3)-(298xx-278)color(white)(x)"kJ/mol")#

#sf(DeltaG^(@)=-181color(white)(x)"kJ/mol")#

#sf((b))#

Average bond enthalpies are usually used to estimate the enthalpy change of a reaction.

In this case we are asked to use the enthalpy change to estimate a bond enthalpy.

Energy needs to be put in to break bonds. Energy is released when new bonds are formed. We can say:

#sf(DeltaH)# = energy put in - energy released.

It helps to draw out the structures involved:

MFDocs

You can see that we need to break 1 N#-=#N bond and 3 F - F bonds.

We then form 6 N - F bonds.

Energy put in = #sf(946+3E(F-F)color(white)(x)kJ)#

Energy released = #sf(6xx272=1632color(white)(x)kJ)#

#sf(DeltaH)# = energy put in - energy released.

#:.##sf(-264=946+3E(F-F)-1632)#

#sf(3E(F-F)=424color(white)(x)kJ)#

#:.##sf(E(F-F)=424/3=141color(white)(x)kJ)#