Question

Consider the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Given that \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\) is 173.4 \(\mathrm{kJ} / \mathrm{mol}\), (a) calculate the standard free energy of formation of \(\mathrm{NO},\) and (b) calculate \(K_{P}\) of the reaction. (c) One of the starting substances in smog formation is NO. Assuming that the temperature in a running automobile engine is \(1100^{\circ} \mathrm{C},\) estimate \(K_{P}\) for the above reaction. (d) As farmers know, lightning helps to produce a better crop. Why?

Short answer

a) The standard free energy of formation for NO is 86.7 kJ/mol. b) The equilibrium constant \(K_{P}\) for the reaction at 25°C is \(4.15*10^{-28}\). c) At 1100°C, \(K_{P}\) is about 179.3. d) Lightning helps produce better crops because it enables the formation of nitrates, a kind of fertilizer.

Step-by-step solution

Calculate the standard free energy of formation for NO

The given balance equation is \(N_{2}(g)+O_{2}(g) \rightleftharpoons 2 NO(g)\). The standard free energy of formation of elemental species in their standard states is zero. So, for \(N_{2}(g)\) and \(O_{2}(g)\), \(\Delta G_f^{\circ}\) is 0. The standard free energy change for the reaction \(\Delta G^{\circ}\) is 173.4 kJ/mol. Using the equation \(\Delta G^{\circ} = \Delta G_f^{\circ} (products) - \Delta G_f^{\circ} (reactants)\), we get \(\Delta G^{\circ} = 2 \Delta G_f^{\circ} (NO(g)) - 0\). Solving this gives \(\Delta G_f^{\circ} (NO(g))\) = 173.4/2 = 86.7 kJ/mol.

Calculate \(K_{P}\) for the reaction

The equation relating standard free energy change and equilibrium constant is given by \(\Delta G^{\circ} = -RT ln(K_{P})\), where R is the gas constant (8.314 J/K*mol) and T is the temperature in Kelvin (25°C = 298 K). Substituting the values, we get \(-173.4*10^3 J = -(8.314 J/K*mol * 298 K) ln(K_{P})\) , solving this yields \(K_{P} = e^{-173.4*10^3/(8.314*298)} = 4.15*10^{-28}\).

Estimate \(K_{P}\) at a different temperature

In order to calculate the equilibrium constant at a different temperature we use the Van't Hoff equation. We can rearrange the equation to: \(ln(K_2/K_1) = - \Delta H_m / R * (1/T_2 - 1/T_1)\), which gives us: \(ln(K_{P2}/K_{P1}) = (-\Delta H_m / 8.314) * (1/1373 - 1/298)\). DeltaH_m is the enthalpy change for the reaction, which can be calculated from DeltaG_m and T: \(\Delta H_m = \Delta G_m + T\Delta S_m = \Delta G_m + T(\Delta G_m / T) = 2*\Delta G_m = 2*173.4 kJ/mol\). Solving the Van't Hoff equation gives: \(K_{P2} = K_{P1} * exp[(-2*173.4*10^3 / 8.314) * (1/1373 - 1/298)] = 4.15 * 10^{-28} * exp[(-2*173.4*10^3 / 8.314) * (1/1373 - 1/298)] = 179.3\).

Explain why lightning helps produce better crops

Lightning helps produce better crops because the electrical energy from lightning can cause nitrogen and oxygen in the atmosphere to react and form nitrogen oxides, including NO and NO2. These nitrogen oxides dissolve in rain to form nitric acid, which is a form of nitrogen that plants can absorb and use to build proteins and other complex molecules. When this 'nitrate' falls to the ground in rain, it gets into the soil and provides plants with necessary nutrients.

Key concepts

Gibbs Free Energy

Gibbs Free Energy is a vital concept in understanding chemical equilibrium, indicating the maximum amount of work that can be performed by a chemical process at constant temperature and pressure. For the reaction \(\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)\), knowing the change in Gibbs Free Energy \(\Delta G^{\circ}\) helps assess whether the reaction is spontaneous. A negative \(\Delta G^{\circ}\) points to a spontaneous process, while a positive value indicates non-spontaneity under standard conditions.

Let's break it down a bit more:

• A reaction with \(\Delta G^{\circ} < 0\) is likely to proceed forward under standard conditions.
• \(\Delta G^{\circ} = 0\) signifies a system at equilibrium.
• For \(\Delta G^{\circ} > 0\), the reaction tends to be non-spontaneous.

In the given exercise, the value \(\Delta G^{\circ} = 173.4\, \mathrm{kJ/mol}\) suggests that the formation of \(\mathrm{NO}\) is not favorable under standard conditions. However, temperature changes can affect this, as reflected by the equilibrium constant \(K_p\) and the temperature-dependency expressed in the Van't Hoff equation.

Equilibrium Constant (Kp)

The Equilibrium Constant \(K_p\) offers insight into the position of equilibrium for gas-phase reactions. It is a measure of the relative concentrations of reactants and products at equilibrium. For a reaction at constant temperature, \(K_p\) remains constant as well.

How \(K_p\) is Related to Gibbs Free Energy:

• The equation \(\Delta G^{\circ} = -RT \ln(K_{p})\) connects \(\Delta G^{\circ}\) directly with \(K_p\), where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin.
• At room temperature (298 K), with \(\Delta G^{\circ} = 173.4\, \mathrm{kJ/mol}\), the \(K_p\) value was calculated as \(4.15 \times 10^{-28}\), indicating reactants prevail.

Because this \(K_p\) is incredibly small, it stresses that at 25°C, the concentration of \(\mathrm{NO}\) is minimal compared to \(\mathrm{N}_2\) and \(\mathrm{O}_2\).
This finding tells chemists about the unfavored nature of \(\mathrm{NO}\) formation under such conditions, an essential factor when considering emissions in car engines or environmental chemistry scenarios.

Van't Hoff Equation

The Van't Hoff Equation provides a way to understand how changes in temperature affect chemical equilibria. It expresses the relationship between the equilibrium constants \(K_p\) at two different temperatures: \(T_1\) and \(T_2\).

The formula is:
\[ \ln\left(\frac{K_{2}}{K_{1}}\right) = \frac{-\Delta H_{m}}{R} \left(\frac{1}{T_{2}} - \frac{1}{T_{1}}\right)\]
Here, \(\Delta H_m\) is the change in enthalpy, a measure of heat absorbed or released.

Using this equation in the exercise, we observe how \(K_p\) at an elevated temperature \(1100^{\circ} \text{C} = 1373 \text{K}\) significantly increases to \(179.3\), from the earlier \(4.15 \times 10^{-28}\) at 25°C.

• This increase reflects that higher temperatures favor the endothermic reaction, producing more \(\mathrm{NO}\).
• Real-world applications include understanding the formation of \(\mathrm{NO}\) in high-temperature environments like car engines, affecting atmospheric chemistry.

Through the Van't Hoff equation, it becomes clear how different thermodynamic drivers shape equilibrium state positions across varying temperatures, providing crucial insight into both chemical industries and environmental implications.