# 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 ?

Nov 7, 2016

$\textsf{\left(a\right)}$

$\textsf{- 181 \textcolor{w h i t e}{x} \text{kJ/mol}}$

$\textsf{\left(b\right)}$

$\textsf{141 \textcolor{w h i t e}{x} \text{kJ}}$

#### Explanation:

$\textsf{\left(a\right)}$

The relationship you need is:

$\textsf{\Delta {G}^{\circ} = \Delta {H}^{\circ} - T \Delta {S}^{\circ}}$

$\therefore$$\textsf{\Delta {G}^{\circ} = - 264 \times {10}^{3} - \left(298 \times - 278\right) \textcolor{w h i t e}{x} \text{kJ/mol}}$

$\textsf{\Delta {G}^{\circ} = - 181 \textcolor{w h i t e}{x} \text{kJ/mol}}$

$\textsf{\left(b\right)}$

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:

$\textsf{\Delta H}$ = energy put in - energy released.

It helps to draw out the structures involved: You can see that we need to break 1 N$\equiv$N bond and 3 F - F bonds.

We then form 6 N - F bonds.

Energy put in = $\textsf{946 + 3 E \left(F - F\right) \textcolor{w h i t e}{x} k J}$

Energy released = $\textsf{6 \times 272 = 1632 \textcolor{w h i t e}{x} k J}$

$\textsf{\Delta H}$ = energy put in - energy released.

$\therefore$$\textsf{- 264 = 946 + 3 E \left(F - F\right) - 1632}$

$\textsf{3 E \left(F - F\right) = 424 \textcolor{w h i t e}{x} k J}$

$\therefore$$\textsf{E \left(F - F\right) = \frac{424}{3} = 141 \textcolor{w h i t e}{x} k J}$