The equilibrium constant, Kc for the reaction, \(NO_{(g)} + \frac{1}{2}O_{2(g)} \rightarrow NO_{2(g)}\), is 35.2. What...
The equilibrium constant, Kc for the reaction, \(NO_{(g)} + \frac{1}{2}O_{2(g)} \rightarrow NO_{2(g)}\), is 35.2. What is the value of K for the reaction, \(NO_{2(g)} \rightarrow NO_{(g)} + \frac{1}{2}O_{2(g)}\)
- A) 35.2
- B) 17.6
- C) 2.84
- D) 1.24
Correct Answer: C) 2.84
Explanation
The given reaction is:
\(NO_{(g)} + \frac{1}{2}O_{2(g)} \rightarrow NO_{2(g)}\)
Its equilibrium constant, Kc, is 35.2. We are asked to find the equilibrium constant for the reverse reaction:
\(NO_{2(g)} \rightarrow NO_{(g)} + \frac{1}{2}O_{2(g)}\)When we reverse a reaction, the new equilibrium constant is the reciprocal of the original equilibrium constant. In other words, if the original reaction has an equilibrium constant Kc, then the reverse reaction has an equilibrium constant 1/Kc.
To find the value of K for the reverse reaction, we simply take the reciprocal of Kc:
\(K = \frac{1}{Kc} = \frac{1}{35.2}\)Calculating this value, we get:
\(K = 2.84 \times 10^{-2}\)So, the value of K for the reverse reaction is 2.84