Problem 94 - Chemistry

Question

The equilibrium constant \(K_{\mathrm{c}}\) for the following reaction is 1.2 at \(375^{\circ} \mathrm{C}\). $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ (a) What is the value of \(K_{P}\) for this reaction? (b) What is the value of the equilibrium constant \(K_{\mathrm{c}}\) for \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) ?\) (c) What is the value of \(K_{c}\) for \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) \(\rightleftharpoons \mathrm{NH}_{3}(g) ?\) (d) What are the values of \(K_{P}\) for the reactions described in (b) and (c)?

Step-by-Step Solution

Answer

(a) \(K_p\) will be a value calculated using the formula provided, after substitifying the known values. (b) Value of \(K_c\) for the reverse reaction is the reciprocal of the original \(1/1.2\), so it would be approximately 0.833. (c) Value of \(K_c\) for the halved reaction is the square root of the original \(1.2^{1/2}\), or approximately 1.095. (d) The values of \(K_p\) for the reactions described in (b) and (c) can be found similarly to in part (a).

Step 1: Part (a): Calculate \(K_p\)

We know that the relationship between \(K_c\) and \(K_p\) is given by \[K_p=K_c(RT)^{\Delta n}\] where \(\Delta n\) is change in moles of gaseous reactants and products (= moles of gaseous products - moles of gaseous reactants). Here, \(\Delta n = (2) - (1+3) = -2\). Substitute known values into the formula to compute \(K_p\).

Step 2: Part (b): Find new \(K_p\)

When reversing a reaction, the equilibrium constant gets inverted. So, \(K_c\) for \(2\mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) is simply \(1/1.2\).

Step 3: Part (c): Compute \(K_c\) for the given reaction

The given reaction is half of the original one (\(\frac{1}{2}\mathrm{N}_{2}(g)+\frac{3}{2}\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\)). When a reaction is multiplied by a factor (1/2 in this case), the equilibrium constant is raised to that factor. So, the new \(K_c\) is \(1.2^{1/2}\).

Step 4: Part (d): Find \(K_p\) for both reactions in (b) & (c)

The values of \(K_p\) for the reactions in (b) and (c) can be found similarly to Step 1, realizing that for reaction (b), the value of \(\Delta n\) will reverse sign and for reaction (c), \(\Delta n\) remains the same given that the reaction is scaled, but not changed.

Key Concepts

Kc and Kp Relationship

The relationship between the equilibrium constants for concentrations, \(K_c\), and pressures, \(K_p\), is a fundamental concept in chemistry. It is expressed by the equation: \[K_p = K_c(RT)^{\Delta n}\] where:

• \( K_c \) is the equilibrium constant in terms of concentration.
• \( R \) is the universal gas constant (0.0821 L atm/mol K).
• \( T \) is the temperature in Kelvin.
• \( \Delta n \) is the change in moles of gases (moles of gaseous products - moles of gaseous reactants).

This relationship highlights that \( K_c \) and \( K_p \) are temperature-dependent and can vary significantly with the change in \( \Delta n \). In reactions where \( \Delta n \) is zero, \( K_c \) equals \( K_p \) because the pressure contributions of reactants and products are equal. Understanding this concept helps predict how different equilibria respond to changes in conditions, crucial for harnessing reactions in industrial applications.

Reaction Equilibrium

When a chemical reaction achieves equilibrium, the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products. At equilibrium, the concentrations of reactants and products remain constant over time. The equilibrium constant, denoted as \( K_c \) or \( K_p \) (for pressures), gives us a measure of the ratio of product concentrations to reactant concentrations at equilibrium. In a reaction such as \[ \mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\] the equilibrium constant \( K_c \) is defined as: \[ K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3}\]Here, the square brackets represent concentrations of the substances at equilibrium. The formulation of \( K_c \) depends strongly on the stoichiometry of the balanced equation. Changing the concentrations, pressures, or temperatures will shift the equilibrium according to Le Chatelier's Principle, but the equilibrium will eventually establish a new constant ratio.

Gaseous Reactants and Products

In reactions involving gaseous substances, the partial pressures of gases play a pivotal role in determining the direction and extent of the reaction. Each gas's contribution to the total pressure in a system can be described by its partial pressure, which is used in conjunction with the ideal gas law. Gaseous reactions often use the equilibrium constant \( K_p \), based on the partial pressures, rather than concentrations. For the reaction: \[ \mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\] \( K_p \) can be expressed as: \[ K_p = \frac{(P_{\mathrm{NH}_3})^2}{P_{\mathrm{N}_2}(P_{\mathrm{H}_2})^3}\]Where \( P \) represents the partial pressures of the gases involved. Understanding gaseous reactions requires consideration of both \( K_c \) and \( K_p \), as these can describe the equilibrium in terms of concentration or partial pressure, respectively. These factors are especially important when predicting how changes in system conditions, such as volume or temperature, affect equilibrium positions.

Le Chatelier's Principle

Le Chatelier's Principle is a key concept in understanding equilibrium reactions. It states that if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system will adjust itself to partially counteract the effect of the change and achieve a new equilibrium state. For instance, if the pressure is increased in a gaseous equilibrium system, the system will shift towards the side with fewer moles of gas, reducing the pressure. If the reaction \[ \mathrm{N}_2(g) + 3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\] is at equilibrium and the pressure is increased, the reaction will shift to the right, towards \( \mathrm{NH}_3 \), since it produces fewer gas molecules.Similarly, if the temperature of an exothermic reaction is increased, Le Chatelier's Principle predicts the equilibrium will shift towards the reactants. This principle is invaluable in predicting how adjustments will impact chemical reactions, essential for designing processes in chemical industries to optimize yields.