Question

Calculate the enthalpy change, \(\Delta \mathrm{H}^{\circ}\), for the reaction $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{NO}(\mathrm{g}) $$ given the equilibrium constants \(4.08 \times 10^{-4}\) for a temperature of \(2000^{\circ} \mathrm{K}\) and \(3.60 \times 10^{-3}\) for a temperature of \(2500^{\circ} \mathrm{K}\).

Short answer

The enthalpy change, \(\Delta \mathrm{H}^{\circ}\), for the given reaction is approximately 158293 J/mol, found by applying the van 't Hoff equation at the given equilibrium constants and temperatures, and solving the equations simultaneously.

Step-by-step solution

Expressing the equilibrium constant in terms of enthalpy and temperature

The van 't Hoff equation shows the relationship between the equilibrium constant (K), the standard enthalpy change of the reaction (\(\Delta \mathrm{H}^{\circ}\)), the temperature (T), and the gas constant (R = 8.314 J/mol K): \( \ln{K} = -\dfrac{\Delta \mathrm{H}^{\circ}}{R} \times \dfrac{1}{T} + C\), where C is the integration constant.

Apply the van 't Hoff equation at given temperatures

Apply the equation at the given equilibrium constants and temperatures. (i) At T = 2000 K and K = \(4.08 \times 10^{-4}\), \( \ln{(4.08 \times 10^{-4})} = -\dfrac{\Delta \mathrm{H}^{\circ}}{8.314} \times \dfrac{1}{2000} + C\) (ii) At T = 2500 K and K = \(3.60 \times 10^{-3}\), \( \ln{(3.60 \times 10^{-3})} = -\dfrac{\Delta \mathrm{H}^{\circ}}{8.314} \times \dfrac{1}{2500} + C\)

Solve the equations simultaneously to find enthalpy change

We need to solve equations (i) and (ii) simultaneously to find the value of \(C\) and \(\Delta \mathrm{H}^{\circ}\). Subtract equation (i) from equation (ii): \( \ln{(3.60 \times 10^{-3})} - \ln{(4.08 \times 10^{-4})} = \dfrac{\Delta \mathrm{H}^{\circ}}{8.314} \times \left( \dfrac{1}{2000} - \dfrac{1}{2500} \right)\) Now, we can solve for \(\Delta \mathrm{H}^{\circ}\): \( \Delta \mathrm{H}^{\circ} = \left( \ln{(3.60 \times 10^{-3})} - \ln{(4.08 \times 10^{-4})} \right) \times \dfrac{8.314}{ \left( \dfrac{1}{2000} - \dfrac{1}{2500} \right)}\) After evaluating the expression, we have: \( \Delta \mathrm{H}^{\circ} \approx 158293\ \text{J/mol}\) The enthalpy change \(\Delta \mathrm{H}^{\circ}\) for the given reaction is approximately 158293 J/mol.

Key concepts

van 't Hoff equation

The van 't Hoff equation is a powerful tool in predicting how a reaction's equilibrium constant (K) changes with temperature. It interrelates the equilibrium constant, temperature, standard reaction enthalpy, and the universal gas constant. This equation is represented as:\[\ln{K} = -\dfrac{\Delta \mathrm{H}^{\circ}}{R} \times \dfrac{1}{T} + C\]where:

• \( K \) = Equilibrium constant
• \( \Delta \mathrm{H}^{\circ} \) = Standard enthalpy change of the reaction
• \( R \) = Universal gas constant (8.314 J/mol K)
• \( T \) = Temperature in Kelvin
• \( C \) = An integration constant

The equation emerges from the relationship between thermal dynamics and equilibrium processes. It implies that the equilibrium constant is temperature-dependent, with the enthalpy change playing a significant role in determining the direction and extent of these changes. As the temperature rises, the equilibrium constant shifts according to the sign and magnitude of \( \Delta \mathrm{H}^{\circ} \), giving vital insights into the feasibility and spontaneity of reactions.When using the van 't Hoff equation, analyzing shifts in \( K \) for varying temperatures helps identify the effect of heat (endothermic or exothermic) on a chemical system.

equilibrium constant

The equilibrium constant (K) is a key concept in chemical reactions, representing the ratio of product to reactant concentrations at equilibrium. For the reaction:\[\mathrm{N}_{2}( ext{g}) + \mathrm{O}_{2}( ext{g}) \rightleftharpoons 2 \mathrm{NO}( ext{g})\]The equilibrium constant expression would be:\[K_c = \dfrac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]}\]Where \( [X] \) denotes the concentration of chemical species X. In gaseous reactions, pressures can sometimes replace concentrations leading to \( K_p \), a pressure-based equilibrium constant.An equilibrium constant provides a snapshot of a chemical system at a given temperature. The value of \( K \) indicates the direction the reaction favors:

• If \( K > 1 \), the reaction favors products, indicating a forward reaction predominates.
• If \( K < 1 \), reactants are favored, suggesting a predominance of the reverse reaction.
• If \( K \approx 1 \), the reaction has a significant amount of both reactants and products.

In the provided exercise, differing values of \( K \) at varying temperatures underline its temperature sensitivity.

reaction enthalpy

Reaction enthalpy or the standard enthalpy change (\( \Delta \mathrm{H}^{\circ} \)) reflects the heat exchange during a chemical reaction. It's defined as the energy change when reactants convert to products under standard conditions (1 atm pressure, 298 K temperature). The units are typically Joules or kilojoules per mole.Enthalpy change carries crucial information:- **Positive \( \Delta \mathrm{H}^{\circ} \):** An endothermic process that absorbs heat. This signifies that heat energy is required to convert reactants into products.
It is often associated with breaking bonds.- **Negative \( \Delta \mathrm{H}^{\circ} \):** An exothermic reaction that releases heat, often linked with forming new bonds.
It implies a spontaneous tendency due to energy release.In the provided calculation, the solved value of \( 158293 \) J/mol indicates a significant endothermic change for the formation of nitrogen monoxide from nitrogen and oxygen. Through understanding \( \Delta \mathrm{H}^{\circ} \), one can gauge the heat dynamics of the reaction and predict how modifying temperatures or reactant conditions might drive or restrain the process.