Question

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction: $$ \mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{I}(g) $$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C}\). Calculate \(K_{c}\) and \(K_{p}\) for the equilibrium $$ 2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g) $$ at the same temperature.

Short answer

For the reverse reaction, \(K_{c} = 2.63 \times 10^{4}\) and \(K_{p} \approx 2159.85\).

Step-by-step solution

Understanding Reversibility of the Reaction

The equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. This means if you are given the equilibrium constant for a reaction, the equilibrium constant for the reverse reaction is simply the inverse of that value.

Calculate Kc for the Reverse Reaction

Given that the equilibrium constant \(K_{c}\) for the forward reaction \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(3.8 \times 10^{-5}\), the \(K_{c}\) for the reverse reaction \(2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g)\) is given by \[K_{c,\text{reverse}} = \frac{1}{3.8 \times 10^{-5}} = 2.63 \times 10^{4}\]

Use the Relationship Between Kc and Kp

For the reaction \(2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g)\), the relationship between \(K_{c}\) and \(K_{p}\) can be expressed using the formula: \[K_{p} = K_{c}(RT)^{\Delta n}\]Where \(R\) is the ideal gas constant (0.08206 L·atm/mol·K), \(T\) is the temperature in Kelvin (727°C = 1000 K), and \(\Delta n\) is the change in moles of gas (\(-1\) in this case since \(2 - 1 = 1\)).

Calculate Kp for the Reverse Reaction

Substitute the known values into the equation: \[K_{p} = (2.63 \times 10^{4})(0.08206 \times 1000)^{-1}\]Calculate the expression:\[K_{p} = (2.63 \times 10^{4})(0.08206)^{-1}\]This results in:\[K_{p} = (2.63 \times 10^{4}) \times (12.177)^{-1} \approx 2159.85\]

Key concepts

Equilibrium Constant (Kc)

In chemistry, the equilibrium constant, denoted as \( K_c \), is a key value that provides insight into the position of equilibrium in a chemical reaction. It can be understood by focusing on the concentrations of the reactants and products at equilibrium. For a general reaction:

• \( aA + bB \rightleftharpoons cC + dD \)

\( K_c \) is calculated using the formula:\[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]where \([A]\), \([B]\), \([C]\), and \([D]\) represent the molar concentrations of the substances involved.
A high \( K_c \) value indicates that the reaction favors products, while a low \( K_c \) suggests a reaction favoring reactants. In the exercise, we inverted the original \( K_c \) because the reaction was reversed, illustrating the concept of reaction reversibility.

Reaction Reversibility

Reversibility in chemical reactions means that the reaction can go in either direction, from reactants to products or the reverse. This feature depends significantly on the conditions such as temperature and pressure.
It's crucial to understand that the equilibrium constant for the reverse reaction is simply the inverse of the original \( K_c \). In our case, for the reaction \( \mathrm{I}_2(g) \rightleftharpoons 2\mathrm{I}(g) \), the given \( K_c \) maintains information about molar concentrations when the reaction proceeds in a specific direction.

• To find the equilibrium constant \( K_{c,\text{reverse}} \) for the reaction \( 2\mathrm{I}(g) \rightleftharpoons \mathrm{I}_2(g) \), we simply calculated the reciprocal, giving us \( 2.63 \times 10^4 \).

This reciprocal relationship is pivotal in understanding how dynamic and adaptable chemical equilibria can be.

Equilibrium Constant (Kp)

The equilibrium constant \( K_p \) is a form of equilibrium constant applicable when dealing with gaseous reactions. It relates to the partial pressures of gases, instead of concentrations. For reactions involving gases, the relationship between \( K_c \) and \( K_p \) is expressed with the equation:\[K_p = K_c(RT)^\Delta n\]where:

• \( R \) is the ideal gas constant (0.08206 L·atm/mol·K).
• \( T \) represents the absolute temperature in Kelvin.
• \( \Delta n \) is the change in moles of gas, calculated by subtracting the moles of gaseous reactants from products.

In our exercise, the \( \Delta n \) was \(-1\). Using the given temperature of 1000 K, we calculated the equilibrium constant \( K_p \) for the reverse reaction as approximately 2159.85. This result exemplifies how equilibrium constants help predict how the reaction conditions will impact the extent to which a reaction occurs under controlled gaseous conditions.