Problem 117 - Chemistry: An Atoms-Focused Approach

Question

Air Pollution Automobiles and trucks pollute the air with NO. At \(2000^{\circ} \mathrm{C}, K_{c}\) for the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ is \(4.10 \times 10^{-4},\) and \(\Delta H^{*}=180.6 \mathrm{kJ} .\) What is the value of \(K_{c}\) at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Answer: The value of \(K_c\) at \(25^{\circ} \mathrm{C}\) is approximately \(4.41 \times 10^{-7}\).

Step 1: Convert the temperature to Kelvin

First, we need to convert the given temperatures from degrees Celsius to Kelvin since we will be working with temperature in the van 't Hoff equation. To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature. So: \(T_1 = 2000^{\circ}\mathrm{C} + 273.15 = 2273.15\ \mathrm{K}\) \(T_2 = 25^{\circ}\mathrm{C} + 273.15 = 298.15\ \mathrm{K}\)

Step 2: Write down the van 't Hoff equation

The van 't Hoff equation is used to relate the temperature dependence of the reaction's equilibrium constants with the enthalpy change of the reaction. The equation is expressed as: \(\frac{-\Delta H^\star}{R} = \frac{\ln K_2 - \ln K_1}{\frac{1}{T_2} - \frac{1}{T_1}}\) Where \(K_1\) and \(K_2\) are the equilibrium constants at temperatures \(T_1\) and \(T_2\), respectively, \(\Delta H^\star\) is the enthalpy change for the reaction, and \(R\) is the universal gas constant which is equal to \(8.314\ \mathrm{J}.\mathrm{mol}^{-1}.\mathrm{K}^{-1}\).

Step 3: Insert the values in the van 't Hoff equation

Now, insert the known values into the van 't Hoff equation: \(\frac{-180.6 \times 10^3 \ \mathrm{J}.\mathrm{mol}^{-1}}{8.314\ \mathrm{J}.\mathrm{mol}^{-1}.\mathrm{K}^{-1}} = \frac{\ln K_2 - \ln 4.10\times10^{-4}}{\frac{1}{298.15\ \mathrm{K}} - \frac{1}{2273.15\ \mathrm{K}}}\)

Step 4: Solve for \(\ln K_2\)

After inserting the values, our primary goal is to find the value of \(\ln K_2\). To solve the equation: \(-2743.28 \approx \frac{\ln K_2 + 8.199}{0.00234}\) Now, multiply both sides by \(0.00234\): \(-6.409 \approx \ln K_2 + 8.199\) Subtract \(8.199\) from both sides: \(\ln K_2 \approx -14.608\)

Step 5: Find \(K_2\) value

Now, to find the value of \(K_2\), take the exponent of both sides: \(K_2 \approx e^{-14.608}\) The value of \(K_2\) is approximately \(4.41 \times 10^{-7}\). So, the value of \(K_c\) at \(25^{\circ} \mathrm{C}\) is approximately \(4.41 \times 10^{-7}\).

Key Concepts

Reaction Kinetics

When we talk about reaction kinetics, we are diving into the study of reaction rates and how different factors influence these rates. Think of reaction kinetics as understanding the speed at which a chemical reaction occurs.
Several factors can impact these rates, such as temperature, concentration, and the presence of catalysts. In the context of the given exercise, the equilibrium constant ( K _c) varies with temperature, which is vital because it directly affects the speed and extent of the reaction converting nitrogen ( N _2) and oxygen ( O _2) into nitric oxide (NO).

• **Temperature:** As a general rule, an increase in temperature usually speeds up chemical reactions. This is because particles have more energy and move faster, resulting in more frequent and forceful collisions.
• **Concentration:** Higher concentrations mean more particles in a given volume, which also leads to more collisions and potentially a faster reaction rate.
• **Catalysts:** Although not applicable in this specific equilibrium exercise, catalysts significantly accelerate reactions without being consumed by offering a lower energy route.

Understanding these kinetics principles helps explain why equilibrium constants change with temperature, as illustrated by the chemical reaction in air pollution scenarios involving NO.

Thermodynamics

Thermodynamics in chemistry revolves around the study of energy changes, particularly the exchange of heat between systems and surroundings. It's all about how energy conversion affects matter in a reaction.
Most importantly, thermodynamics helps us understand which reactions are energetically feasible and how conditions like temperature impact these reactions. The exercise focuses on the enthalpy change (\(\Delta H^*\)) for the reaction converting N_2 and O_2 into NO, which is given as 180.6 kJ/mol. This positive value indicates the reaction is endothermic, meaning it absorbs heat from its surroundings.

• **Enthalpy (\(\Delta H^*\))**: Represents the heat change at constant pressure. A positive \(\Delta H^*\) signifies an endothermic process, absorbing heat.
• **Entropy and Gibbs Free Energy**: Although not detailed in this particular exercise, these concepts are crucial for determining reaction spontaneity along with enthalpy.
• **Energy Conservation**: The first law of thermodynamics, or the principle of energy conservation, implies that energy in a system is constant, merely changing forms.

In our problem, thermodynamics aids in determining how equilibrium predictions made at 2000°C alter when shifted to 25°C, thanks to the Van 't Hoff Equation.

Van 't Hoff Equation

The Van 't Hoff Equation is a pivotal principle in thermodynamics used to calculate how the equilibrium constant of a reaction (K_c) varies with temperature. It interlinks reaction kinetics and thermodynamics, bridging the concept of how heat changes impact reaction scorings.
The equation used in the exercise:\[\frac{-\Delta H^*}{R} = \frac{\ln K_2 - \ln K_1}{\frac{1}{T_2} - \frac{1}{T_1}}\]

• **\(\Delta H^*\)**: Refers to the enthalpy change and is expressed in Joules per mole (J/mol).
• **\(R\)**: Universal gas constant, 8.314 J/(mol K), a fundamental constant used in many chemical calculations.
• **Equilibrium Constants (K_1 and K_2)**: Represent the reaction state at initial and final temperatures, showing the balance point of reactants and products.
• **Temperature Change**: Noted here as conversion of Celsius to Kelvin, crucial for accurate results.

Using the Van 't Hoff Equation, we can compute the expected shift in reaction conditions as temperature decreases dramatically from 2000°C to 25°C. This gives students insights into how interventions like pollution controls can modify the dynamics of chemical equilibria in atmospheric chemistry.