Question

using any data you can find in the aleks data resource, calculate the equilibrium constant k at 25.0 °c for the following reaction.
n₂(g) + o₂(g) → 2 no (g)
round your answer to 2 significant digits.
k =
thermodynamic properties of pure substances
nh₄no₃(s) -365.6 -183.9 151.1
nf₃(g) -132.1 -90.6 260.8
hno₂(g) -79.5 -46.0 254.1
hno₂(aq) -118.8 -53.6 254.1
hno₃(l) -174.1 -80.7 155.6
hno₃(g) -133.9 -73.5 266.9
hno₃(aq) -207.4 -111.3 146.4
n₂o(g) 81.6 103.7 220.0
no(g) 91.3 87.6 210.8
no₂(g) 33.2 51.3 240.1
n₂o₃(g) 86.6 142.4 314.7
n₂o₄(g) 11.1 99.8 304.4
n₂o₅(s) -43.1 113.9 178.2
n₂o₅(g) 13.3 117.1 355.7
no₃⁻(aq) -207.4 -111.3 146.4
nocl(g) 51.7 66.1 261.7
oxygen
o₂(g) 0 0 205.2
o(g) 249.2 231.7 161.1
o₃(g) 142.7 163.2 238.9

Explanation:

Step1: Find ΔG° for the reaction

The formula for ΔG° of a reaction is \( \Delta G^\circ = \sum n\Delta G_f^\circ(\text{products}) - \sum m\Delta G_f^\circ(\text{reactants}) \). For the reaction \( \text{N}_2(\text{g}) + \text{O}_2(\text{g})
ightarrow 2\text{NO}(\text{g}) \), \( \Delta G_f^\circ(\text{N}_2(\text{g})) = 0 \), \( \Delta G_f^\circ(\text{O}_2(\text{g})) = 0 \), and \( \Delta G_f^\circ(\text{NO}(\text{g})) = 87.6 \, \text{kJ/mol} \). So:
\( \Delta G^\circ = 2\times87.6 - (0 + 0) = 175.2 \, \text{kJ/mol} = 175200 \, \text{J/mol} \)

Step2: Relate ΔG° to K using \( \Delta G^\circ = -RT\ln K \)

Rearrange to solve for \( K \): \( \ln K = -\frac{\Delta G^\circ}{RT} \). \( R = 8.314 \, \text{J/(mol·K)} \), \( T = 25.0^\circ\text{C} + 273.15 = 298.15 \, \text{K} \).
Substitute values: \( \ln K = -\frac{175200}{8.314\times298.15} \approx -70.7 \)

Step3: Calculate K

Exponentiate both sides: \( K = e^{\ln K} = e^{-70.7} \approx 4\times10^{-31} \) (rounded to 2 significant digits)

Answer:

\( 4\times10^{-31} \)