Chapter 5: Problem 5

Find the relation between \(K_{p}\) and \(K_{c}\) for the following gas-phase reaction: \(a \mathrm{~A}+b \mathrm{~B} \rightleftarrows c \mathrm{C}+d \mathrm{D}\)

Short Answer

Expert verified

Answer: The relationship between Kp and Kc for a gas-phase reaction is given by the equation \(K_p = K_c(RT)^{(\Delta{n})}\) where \(\Delta{n} = (c+d)-(a+b)\) is the change in the number of gas-phase moles in the reaction and R and T are the gas constant and temperature, respectively.

Step by step solution

Write the expression for the equilibrium constant in terms of concentration (Kc)

The equilibrium constant, \(K_c\) can be written for the given reaction as: \[K_c = \frac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b}\]

Write the expression for the equilibrium constant in terms of pressure (Kp)

Similarly, the equilibrium constant in terms of partial pressures, \(K_p\), can be written as: \[K_p = \frac{(\mathrm{P_C})^c (\mathrm{P_D})^d}{(\mathrm{P_A})^a (\mathrm{P_B})^b}\]

Use the ideal gas law to convert between concentration and pressure

For each species, concentrations can be related to partial pressures through the ideal gas law, \(PV = nRT\) where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature. The formula can be modified to relate concentration and pressure as \[\frac{n}{V} = \frac{P}{RT}\] For each species X in the reaction, we have, \[\frac{[\mathrm{X}]}{RT} = \mathrm{P_X}\]

Re-write Kp in terms of Kc using the above relationship

Substitute the expressions for \(\mathrm{P_A, P_B, P_C, P_D}\) in terms of concentrations into the expression for \(K_p\). The result is: \[ K_p=\frac{\left(\frac{[\mathrm{C}]}{RT}\right)^c \left(\frac{[\mathrm{D}]}{RT}\right)^d} {\left(\frac{[\mathrm{A}]}{RT}\right)^a \left(\frac{[\mathrm{B}]}{RT}\right)^b}\]

Simplify the expression for Kp in terms of Kc

Rearrange and simplify the previous equation: \[K_p = \frac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b}\times\frac{(RT)^{c+d}}{(RT)^{a+b}}\] Since the expression inside the brackets is \(K_c\), we define the relationship between \(K_p\) and \(K_c\) as follows: \[K_p = K_c(RT)^{(c+d)-(a+b)}\] or \[K_p = K_c(RT)^{(\Delta{n})}\] where \(\Delta{n} = (c+d)-(a+b)\) is the change in the number of gas-phase moles in the reaction.

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Find the relation between \(K_{p}\) and \(K_{c}\) for the following gas-phase reaction: \(a \mathrm{~A}+b \mathrm{~B} \rightleftarrows c \mathrm{C}+d \mathrm{D}\)

Short Answer

Step by step solution

Write the expression for the equilibrium constant in terms of concentration (Kc)

Write the expression for the equilibrium constant in terms of pressure (Kp)

Use the ideal gas law to convert between concentration and pressure

Re-write Kp in terms of Kc using the above relationship

Simplify the expression for Kp in terms of Kc

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