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Why is $K_c$ not affected by change in pressure? I know the mathematical explanation, but I don't really understand the reason when only looking at $K_c$. The explanation I know is with the reference to $K_p$, but is it possible to explain without $K_p$?

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    $\begingroup$ Kc and Kp are.constants for a reaction. They dont get affected by changing volume or pressure. They tell you the value of Q at the equilibrium condition. $\endgroup$ – user1825567 Apr 9 '16 at 12:18
  • $\begingroup$ I agree. This sounds like K vs. Q confusion. When the pressure, volume, concentrations, whatever change, then Q changes. However, only one value of Q defines equilibrium. The system responds to reestablish equilibrium to that $Q=K$ once again. $\endgroup$ – Ben Norris Apr 10 '16 at 10:43
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Equilibrium is reached when the forward and reverse reaction rates are equal. These rates are determined by the frequency of successful collisions, which is turn is determined by concentration (seen in the rate laws), temperature, activation energy, and how often molecules collide with the correct orientation (together these last three determine the rate constant, $k$, as shown in the Arrhenius equation).

The equilibrium constant, $K_c$ is the ratio of the rate constants, so only variables that affect the rate constants can affect $K_c$. Pressure doesn't show in any of these relationships.

$$K_c=\frac{k_\text{forward}}{k_\text{reverse}}=\frac{[\text{products}]^y}{[\text{reactants}]^x}$$

$$\text{rate}_\text{forward}=k_\text{forward}[\text{reactants}]^x$$

$$\text{rate}_\text{reverse}=k_\text{reverse}[\text{products}]^y$$

$$k=A\mathrm e^{-\frac{E_\mathrm a}{RT}}$$

It may also be useful to think about different ways pressure can be changed.

  1. The volume of the reaction can be changed. This will affect concentrations of reactants and products, and the reaction will likely have to shift left or right to reestablish equilibrium. If volume were decreased, the rates of the forward and reverse reaction will each increase due to higher concentration. If the exponents in the rate laws are different, the amount of increase for each will be different. Although the rates change, the rate constants do not, so $K_c$ doesn't change either.

  2. The temperature can be changed. This will affect the rate constants of the forward and reverse reactions according to the Arrhenius equation. An increase in temperature will increase $k_\text{forward}$ and $k_\text{reverse}$, but by different amounts because their activation energies are rarely equal. This changes $K_c$, which is the ratio of the two rate constants.

  3. More gas can be added without changing the volume of the container.

    a. If the gas is one of the reactants or products, this would affect one of the concentrations, and the reaction will have to shift to reestablish equilibrium. Again, this changes one of the rates, but does not affect the rate constants, so $K_c$ is unaffected.

    b. If the gas is inert and doesn't react with anything involved in the equilibrium, the concentration of each reactant is unchanged. Neither the rates nor the rate constants are affected, so the system should still be at equilibrium and $K_c$ is unaffected.

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