Problem 111 - Chemistry: An Atoms-Focused Approach

Question

Use the \(\Delta G^{*}\) data given here to calculate the value of \(K_{\mathrm{p}}\) at \(298 \mathrm{K}\) for the following reaction: $$ \begin{array}{c} \mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \\\ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad \Delta G^{*}=173.2 \mathrm{kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta G^{*}=-69.7 \mathrm{kJ} \end{array} $$

Step-by-Step Solution

Answer

Based on the given information, when the reaction N₂(g) + 2O₂(g) ⇌ 2NO₂(g) is at equilibrium at 298 K, the equilibrium constant Kp is approximately 1.95 × 10⁻¹⁸.

Step 1: Determine the Overall Reaction Gibbs Free Energy Change

First, let's add the given reactions to find the net reaction: $$\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g) \quad \Delta G^{*1} = 173.2\,\mathrm{kJ}$$ $$2\mathrm{NO}(g)+\mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}_2(g) \quad \Delta G^{*2} = -69.7\,\mathrm{kJ}$$ Summing the reactions: $$\mathrm{N}_2(g)+2\mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}_2(g)$$ To find the overall \(\Delta G^{*}\), we can sum the given \(\Delta G^{*}\) values: $$\Delta G^{*} = \Delta G^{*1} + \Delta G^{*2} = 173.2\,\mathrm{kJ} + (-69.7\,\mathrm{kJ}) = 103.5\,\mathrm{kJ}$$

Step 2: Calculate the Equilibrium Constant Kp

Now that we have the overall \(\Delta G^{*}\), we can use the relationship between \(\Delta G^{*}\) and \(K_p\) to find Kp. The equation is: $$\Delta G^{*} = -RT \ln K_p$$ Rearrange the equation to solve for \(K_p\): $$K_p = e^{-\frac{\Delta G^{*}}{RT}}$$ Plug in the values for \(\Delta G^{*}\), \(R\), and \(T\): $$K_p = e^{-\frac{103500\,\mathrm{J/mol}}{(8.314\,\mathrm{J/(mol·K)})(298\,\mathrm{K})}}$$ Finally, calculate the value of \(K_p\): $$K_p = 1.95 \times 10^{-18}$$ So, the value of \(K_p\) for the given reaction is approximately \(1.95 \times 10^{-18}\) at 298 K.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G\)) helps us understand the spontaneity of chemical reactions. It combines both enthalpy and entropy, captured in the formula:\[\Delta G = \Delta H - T\Delta S\]where \(\Delta H\) represents change in enthalpy, \(T\) temperature, and \(\Delta S\) change in entropy. A reaction is spontaneous when \(\Delta G\) is negative. In the exercise, we're given \(\Delta G^{*}\) values for two reactions. By summing these values, we get the overall \(\Delta G^{*}\) for the formation of \(\mathrm{NO}_2\) from \(\mathrm{N}_2\) and \(\mathrm{O}_2\). This cumulative energy change indicator serves as a bridge to calculate the equilibrium constant \(\mathrm{K}_p\), connecting thermodynamic and kinetic understandings in one formula.

Thermodynamics

Thermodynamics more broadly considers how heat and energy flow in chemical processes. It examines how they influence reaction spontaneity and equilibrium. The First Law (conservation of energy) and Second Law of Thermodynamics (entropy increases in spontaneous processes) frame chemical transformations. In this exercise:

• We observed the summation principle, using individual \(\Delta G^{*}\) values to deduce the net reaction's energy characteristics.
• By calculating \(\Delta G^{*}\) for the net reaction, we applied the concept of energetics on a macroscopic thermodynamic plane to evaluate the reaction pathway.

These principles allow us to predict whether a reaction will proceed under certain conditions, crucial for practical applications in fields like chemistry and engineering.

Chemical Equilibrium

Chemical Equilibrium occurs when the forward and reverse reaction rates are identical, maintaining constant concentrations of reactants and products. The equilibrium constant (\(K\)) helps describe these scenarios, with large values indicating products dominate at equilibrium and small values suggesting reactants are more prevalent. In equilibrium:

• The reaction can shift left or right in response to changes in conditions, described by Le Chatelier’s principle.
• In this problem, \(K_p\) describes the pressure-based equilibrium state at 298 K.

Using the formula \(\Delta G^{*} = -RT \ln K_p\), we connect \(\Delta G^{*}\) to equilibrium. Here, we calculated a very small \(K_p\), suggesting that at equilibrium, reactants dominate this reaction.

Reaction Kinetics

Reaction Kinetics is about the speed of reactions - how fast reactants convert into products. It relates to the mechanism and pathway which reactions follow, distinct from equilibrium concepts which only tell about final states. While \(\Delta G\) and \(K\) describe the thermodynamics of reactions, kinetics informs about rate and stepwise progression under varying conditions like temperature.

• Kinetics does not affect equilibrium position but explains how quickly it is reached.
• The \(K_p\) value derived shows a high energy barrier; thus, even if conditions become favorable, the reaction may proceed slowly.

Thus, both kinetics and equilibrium offer complementary views of reactions, harmonizing to fully describe chemical processes in scientific and industrial contexts.