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 given reaction. (d) As farmers know, lightning helps to produce a better crop. Why?

Short answer

(a) 86.7 kJ/mol (b) \(3.12 \times 10^{-31}\) (c) \(2.77 \times 10^{-7}\) (d) Lightning aids in nitrogen fixation.

Step-by-step solution

Understand the given reaction

The chemical equation is given as \( \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \). The change in the Gibbs free energy for this reaction at 25°C is provided as \( \Delta G^{\circ} = 173.4 \; \mathrm{kJ/mol} \). Our goal is to use this information to solve additional related thermodynamic properties.

Calculate the standard free energy of formation of NO

The equation for the standard Gibbs free energy change is: \( \Delta G^{\circ}_{reaction} = \sum \Delta G^{\circ}_{f, \text{products}} - \sum \Delta G^{\circ}_{f, \text{reactants}} \). Given that \( \Delta G^{\circ}_{f, \mathrm{N_2}} = 0 \) and \( \Delta G^{\circ}_{f, \mathrm{O_2}} = 0 \), we find:\[\Delta G^{\circ}_{reaction} = 2 \Delta G^{\circ}_{f, \mathrm{NO}} - \left( \Delta G^{\circ}_{f, \mathrm{N_2}} + \Delta G^{\circ}_{f, \mathrm{O_2}} \right)\]\[173.4 = 2 \Delta G^{\circ}_{f, \mathrm{NO}}\]Solving for \( \Delta G^{\circ}_{f, \mathrm{NO}} \):\[\Delta G^{\circ}_{f, \mathrm{NO}} = \frac{173.4}{2} = 86.7 \; \mathrm{kJ/mol}\]

Calculate \(K_{P}\) at 25°C

The relationship between \( \Delta G^{\circ} \) and \( K_P \) is given by the equation: \[\Delta G^{\circ} = -RT \ln K_P\]Where \( R = 8.314 \; \mathrm{J/(mol \cdot K)} \) and \( T = 298 \; \mathrm{K} \). Plug the values into the equation:\[173400 = -(8.314)(298) \ln K_P\]Solving for \( \ln K_P \):\[\ln K_P = -\frac{173400}{8.314 \times 298} \approx -70.01\]Convert \( \ln K_P \) to \( K_P \):\[K_P = e^{-70.01} \approx 3.12 \times 10^{-31}\]

Estimate \(K_{P}\) at 1100°C

First, convert the temperature to Kelvin: \( 1100°C + 273 = 1373 \; \mathrm{K} \). Since \( \Delta G^{\circ} = -RT \ln K_P \), solve for \( K_P \) at 1373 K:\[\ln K_P = -\frac{173400}{8.314 \times 1373} \approx -15.09\]Convert \( \ln K_P \) to \( K_P \):\[K_P = e^{-15.09} \approx 2.77 \times 10^{-7}\]

Explain why lightning aids crop production

Lightning provides the energy to convert \( N_2 \) in the atmosphere to \( NO \), which ultimately forms nitrites and nitrates, essential nutrients for plant growth. This natural process increases the availability of nitrogen in the soil, enhancing crop yield.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental principle in chemistry that describes a state where the rates of the forward and reverse reactions are equal. As a result, the concentrations of all reactants and products remain constant over time. Understanding chemical equilibrium helps us predict the concentrations of different species in a reaction and is crucial for reactions that occur in closed systems.
For instance, in the reaction involving \( \mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g) \), the formation of nitrogen monoxide (\(\mathrm{NO}\)) from nitrogen (\(\mathrm{N_2}\)) and oxygen (\(\mathrm{O_2}\)) is reversible. Once equilibrium is reached, the reaction doesn't stop; instead, the forward reaction rate equals the backward reaction rate, maintaining constant concentrations of \(\mathrm{N_2}\), \(\mathrm{O_2}\), and \(\mathrm{NO}\).
This concept is vital for understanding how various conditions, such as temperature and pressure, affect the outcome of chemical reactions.

Thermodynamics

Thermodynamics is the branch of science concerned with heat and temperature and their relation to energy and work. In reactions like the formation of nitric oxide, thermodynamic principles help determine the feasibility and spontaneity of the reaction.
The Gibbs Free Energy (\(\Delta G\)) is central to thermodynamics in chemical reactions. It combines the concepts of enthalpy (\(\Delta H\)), entropy (\(\Delta S\)), and temperature (\(T\)) in the formula:\[\Delta G = \Delta H - T \Delta S \]A negative \(\Delta G\) implies a spontaneous reaction under constant temperature and pressure. In the given exercise, the \(\Delta G^{\circ}\) value at 25°C is positive, indicating the reaction is non-spontaneous at this temperature.
This relationship emphasizes how changes in thermodynamic properties alter the direction and extent to which reactions occur.

Standard Free Energy of Formation

Standard free energy of formation (\(\Delta G^{\circ}_{f}\)) is the change in Gibbs free energy when one mole of a compound forms from its elements in their standard states. It provides a measure of stability and reactivity of substances.
In the context of nitric oxide (\(\mathrm{NO}\)), we use the formula:\[\Delta G^{\circ}_{reaction} = \sum \Delta G^{\circ}_{f, \text{products}} - \sum \Delta G^{\circ}_{f, \text{reactants}} \]For \(\mathrm{NO}\), both \(\Delta G^{\circ}_{f, \mathrm{N_2}}\) and \(\Delta G^{\circ}_{f, \mathrm{O_2}}\) are zero since they are elemental gases. Given the \(\Delta G^{\circ}_{reaction}\) of 173.4 kJ/mol for the reaction, the \(\Delta G^{\circ}_{f, \mathrm{NO}}\) comes out to 86.7 kJ/mol. This calculated value illustrates the energy required for forming \(\mathrm{NO}\) from its elements and points to the energetic favorability of its formation.

Equilibrium Constant Kp

The equilibrium constant \(K_P\) is a measure of the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their coefficients in the balanced equation. For reactions involving gases, \(K_P\) relates to the partial pressures of the gases.
The relationship between \(\Delta G^{\circ}\) and \(K_P\) is crucial for understanding reaction dynamics:\[\Delta G^{\circ} = -RT \ln K_P\]Where \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. A very small \(K_P\) value, like at 25°C for the reaction forming \(\mathrm{NO}\), suggests negligible products at equilibrium, reinforcing how \(\Delta G^{\circ}\) reflects the reaction's non-spontaneous nature. At higher temperatures, such as 1100°C, \(K_P\) is significantly larger, indicating a shift towards more product formation due to the increased energy available to the system.
Understanding \(K_P\) helps predict how changes in conditions impact the concentrations of reactants and products.