Question

Derive the relationship $$ K=K_{\mathrm{c}} \times(R T)^{\Delta n_{\mathrm{g}}} $$ where \(K_{\mathrm{c}}\) is the equilibrium constant using molarities and \(\Delta n_{\mathrm{g}}\) is the change in the number of moles of gas in the reaction (see page 310 ). (Hint: Recall that \(P_{\mathrm{A}}=n_{\mathrm{A}} R \mathrm{~T} / \mathrm{V}\) and \(\left.n_{\mathrm{A}} / V=[\mathrm{A}] .\right)\)

Short answer

Question: Derive the relationship between the equilibrium constant in terms of molarities (K_c) and the equilibrium constant in terms of partial pressures (K) for an ideal gas. Answer: The relationship between K_c and K for an ideal gas can be derived as: $$K=K_c \times(R T)^{\Delta n_g}$$, where R is the gas constant, T is the temperature in kelvin, and Δn_g is the change in the number of moles of gas in the reaction.

Step-by-step solution

Write the expression for \(K_c\) and \(K\)

Write the expressions for the equilibrium constant \(K_c\) in terms of molar concentration and for the equilibrium constant \(K\) in terms of partial pressures, respectively. For a general reaction: $$aA + bB \rightleftharpoons cC + dD$$ where the lowercase letters represent the stoichiometric coefficients and the uppercase letters represent the species, the expressions for \(K_c\) and \(K\) are: $$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$ $$K = \frac{P_C^c P_D^d}{P_A^a P_B^b}$$

Recall the ideal gas law equation

Recall the equation relating an ideal gas's pressure, moles, temperature, and volume: $$P_A = \frac{n_A R T}{V}$$. Here, \(P_A\) is the partial pressure of species \(A\), \(n_A\) is the number of moles of species \(A\), \(R\) is the gas constant, \(T\) is the temperature in kelvin, and \(V\) is the volume.

Express partial pressures in terms of molar concentrations

Use the ideal gas law equation to express the partial pressures of the species in terms of their molar concentrations by isolating the molar concentration term (\(\frac{n_A}{V}\)): $$\frac{n_A}{V} = \frac{P_A}{R T}$$ Substitute this expression into the \(K\) expression: $$K = \frac{\left(\frac{P_C}{R T}\right)^c\left(\frac{P_D}{R T}\right)^d}{\left(\frac{P_A}{R T}\right)^a\left(\frac{P_B}{R T}\right)^b}$$

Simplify the equation

Rewrite the equation by combining similar terms and instances of \(R T\): $$ K = \frac{[C]^c [D]^d R^{c+d} T^{c+d}}{[A]^a [B]^b R^{a+b} T^{a+b}} $$ Simplify further by cancelling out the similar terms in the numerator and denominator: $$ K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \times \frac{T^{c+d}}{T^{a+b}} $$ $$ K = K_c \times \frac{T^{c+d}}{T^{a+b}} $$

Express the final equation using \(\Delta n_g\)

Use the change in the number of moles of gas in the reaction, \(\Delta n_g = (c + d) - (a + b)\), to rewrite the equation as: $$ K = K_c \times (R T)^{\Delta n_g} $$ The derived relationship is: $$K=K_c \times(R T)^{\Delta n_g}$$