N2(g) + 3 F2(g) → 2 NF3(g) ΔH° = –264 kJ/mol ΔS° = –278 J/(mol∙K)
a. calculate the maximum amount of non-PΔV work that can be accomplished through this reaction at a temperature of 500°C.
b. Determine the temperature at which the equilibrium constant for this reaction is equal to 1.00.
c. Use the information above and the table of average bond enthalpies below to calculate the average bond enthalpy of the F–F bond. (You may find it helpful to draw Lewis structures…)
Bond Average Bond Energy (kJ mol-1)
N≡N 946
N–F 272
F–F ?
141
a. To calculate the maximum amount of non-PΔV work that can be accomplished through this reaction at a temperature of 500°C, you need to use the equation:
ΔG° = ΔH° - TΔS°
Where:
ΔG° is the standard Gibbs free energy change,
ΔH° is the standard enthalpy change, and
ΔS° is the standard entropy change.
Since you have the values for ΔH° and ΔS°, you can substitute them into the equation:
ΔG° = -264 kJ/mol - (500 + 273) K * (-278 J/(mol∙K))
Note that the temperature needs to be in Kelvin (K) for the calculation.
b. To determine the temperature at which the equilibrium constant for this reaction is equal to 1.00, you can use the equation:
ΔG° = -RTln(K)
Where:
ΔG° is the standard Gibbs free energy change,
R is the gas constant (8.314 J/(mol∙K)),
T is the temperature in Kelvin,
ln(K) is the natural logarithm of the equilibrium constant.
Since you want to find the temperature at which K = 1.00, you can rearrange the equation:
ln(K) = -ΔG°/RT
Then, substitute the values:
ln(1.00) = -(-264 kJ/mol)/(8.314 J/(mol∙K) * T)
Simplify and solve for T:
ln(1) = 264000 J/(8.314 J/K/mol * T)
0 = 264000 J/(8.314 J/(K/mol) * T)
T = 264000 J/(8.314 J/(K/mol) * 0
T = 0 K
Therefore, the equilibrium constant for this reaction is equal to 1.00 at 0 Kelvin.
c. To calculate the average bond enthalpy of the F-F bond, you can use the average bond energies given in the table. The bond enthalpy of a bond can be calculated by subtracting the sum of the bond energies of the reactant bonds from the sum of the bond energies of the product bonds.
Given the balanced equation: N2(g) + 3 F2(g) → 2 NF3(g)
The reactant bonds are:
N≡N (1 bond with energy 946 kJ/mol)
F-F (3 bonds with unknown energy)
The product bonds are:
N-F (6 bonds with energy 272 kJ/mol)
To calculate the average bond enthalpy of the F-F bond, you can use the equation:
F-F bond enthalpy = (Sum of reactant bond energies) - (Sum of product bond energies) / Number of bonds
F-F bond enthalpy = (3 * ?) - (6 * 272 kJ/mol) / 3
To solve for the unknown F-F bond energy, you need to rearrange the equation:
3 * ? = 6 * 272 kJ/mol + F-F bond enthalpy * 3
? = (6 * 272 kJ/mol + F-F bond enthalpy * 3) / 3
Now, substitute the values given:
? = (6 * 272 kJ/mol + F-F bond enthalpy * 3) / 3
Simplify and solve for F-F bond enthalpy.