Question

Determine values of \(K_{c}\) from the \(K_{p}\) values given. (a) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=2.9 \times 10^{-2} \mathrm{at} 303 \mathrm{K}\) (b) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\) \(K_{\mathrm{p}}=1.48 \times 10^{4} \mathrm{at} 184^{\circ} \mathrm{C}\) (c) \(\mathrm{Sb}_{2} \mathrm{S}_{3}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Sb}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(K_{\mathrm{p}}=0.429\) at \(713 \mathrm{K}\)

Short answer

The values of \(K_{c}\) for the reactions are (a) 0.011, (b) \(2.2 \times 10^{5}\), and (c) 0.429.

Step-by-step solution

Solution to part (a)

For the equation \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\), \(\Delta n = 2 - 1 = 1\). Therefore, \(K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{2.9 \times 10^{-2}}{(0.0821 \times 303)^1} = 0.011\).

Solution to part (b)

For the equation \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\), \(\Delta n = 2 - 3 = -1\). Also, the temperature needs to be converted from Celsius to Kelvin, thus \(T = 184 + 273 = 457K\). Therefore, \(K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{1.48 \times 10^{4}}{(0.0821 \times 457)^{-1}} = 2.2 \times 10^{5}\).

Solution to part (c)

For the equation \(\mathrm{Sb}_{2} \mathrm{S}_{3}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Sb}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\), \(\Delta n = 3 - 3 = 0\). Therefore, \(K_{p}= K_c\) as \((RT)^{\Delta n} = 1\), hence \(K_{c} = 0.429\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible reaction where the forward and backward reactions occur at the same rate. This means that the concentrations of reactants and products remain constant over time. In equations, chemical equilibrium is often represented by the equilibrium constant, either as \( K_c \) for concentrations or \( K_p \) for partial pressures, especially in gas reactions.
It's important to note that equilibrium does not mean that the reactants and products are equal in concentration. It simply implies that their rates of formation are equal, making the system stable. Understanding equilibrium is crucial in predicting how changes in conditions like temperature or pressure affect a system. By manipulating these factors, you can shift the equilibrium in favor of either reactants or products, known as Le Chatelier’s Principle.

Gas Phase Reactions

Gas phase reactions involve chemical processes where all reactants and products are in the gaseous state. They are significant because gases freely mix and expand, making the reactions often quicker than those in other phases.
The behavior of gases is described by the ideal gas law, \( PV = nRT \), where pressure \( P \), volume \( V \), moles \( n \), the gas constant \( R \), and temperature \( T \) are key components. Understanding these principles helps in calculating changes during reactions like pressure and volume shifts.

• When converting \( K_p \) (pressure-based constant) to \( K_c \) (concentration-based constant) in gas reactions, the change in the number of moles \( \Delta n \) determines the conversion factor \((RT)^{\Delta n}\).

Converting between these constants allows chemists to better understand and predict the outcomes of gas reactions under different conditions.

Thermodynamics in Chemistry

Thermodynamics in chemistry deals with energy changes in chemical reactions. It helps explain why reactions occur and whether they will proceed spontaneously.

• One key thermodynamic term is the free energy change, \( \Delta G \), indicating the spontaneity of a reaction.
• Another is enthalpy, \( \Delta H \), representing heat exchange, and entropy, \( \Delta S \), representing disorder.

In gas reactions, thermodynamics also involves understanding how temperature and pressure affect the equilibrium state.
The relationship between \( K_p \) and \( K_c \) highlights the thermodynamic principles at play. As temperature influences these constants, adjusting temperature can lead to different thermodynamically favored conditions. This makes thermodynamics an essential tool for chemists in designing and controlling chemical processes efficiently.