How is the equilibrium constant for the forward reaction of an equilibrium (Kf) related to...
How is the equilibrium constant for the forward reaction of an equilibrium (Kf) related to that of the reverse reaction (Kr)?
- A) (Kr) is the additive inverse of (Kf)
- B) (Kr) is the multiplicative inverse of (Kf)
- C) (Kr) is the same as (Kf)
- D) the product of (Kr) and (Kf) is zero
Correct Answer: B) (Kr) is the multiplicative inverse of (Kf)
Explanation
In a chemical equilibrium, the forward reaction and the reverse reaction occur at the same rate. The equilibrium constant (K) is a measure of the ratio of the concentrations of products to reactants at equilibrium. The equilibrium constant for the forward reaction (K\(_f\) and the reverse reaction (K\(_r\) are related to each other.
The correct answer isOption B, which states that K\(_r\) is the multiplicative inverse of K\(_f\). This means that the product of the equilibrium constants for the forward and reverse reactions is equal to 1:
\[K_f) \times (K_r) = 1\]To better understand this, let's consider a simple chemical equilibrium:
\[\text{A} \rightleftharpoons \text{B}\]The forward reaction has an equilibrium constant, K\(_f\), and the reverse reaction has an equilibrium constant, K\(_r\). According to the law of mass action, we can express these constants as:
\[K_f = \frac{[\text{B}]}{[\text{A}]}\] \[K_r = \frac{[\text{A}]}{[\text{B}]}\]Now, if we multiply these two equations, we get:
\[K_f) \times (K_r) = \frac{[\text{B}]}{[\text{A}]} \times \frac{[\text{A}]}{[\text{B}]} = 1\]This demonstrates that the equilibrium constants for the forward and reverse reactions are indeed multiplicative inverses of each other.