Question

The equilibrium constant for the given reaction is \(100\).
\[ N_2(g) + 2O_2(g) \rightleftharpoons 2NO_2(g) \]
What is the equilibrium constant for the reaction given below?
\[ NO_2(g) \rightleftharpoons \frac{1}{2}N_2(g) + O_2(g) \]

• A. \(10\)
• B. \(1\)
• C. \(0.1\)
• D. \(0.01\)

Hint:

If reaction is reversed, \(K\) becomes \(1/K\). If coefficients are multiplied by \(n\), new \(K = K^n\).

Answer 1

The Correct Option is C

Step 1: Given equilibrium constant.
For reaction:
\[ N_2 + 2O_2 \rightleftharpoons 2NO_2,\quad K = 100 \]
Step 2: Reverse the reaction.
Reversing gives:
\[ 2NO_2 \rightleftharpoons N_2 + 2O_2 \]
So new constant becomes:
\[ K' = \frac{1}{100} \]
Step 3: Divide the reaction by 2.
Required reaction is:
\[ NO_2 \rightleftharpoons \frac{1}{2}N_2 + O_2 \]
So equilibrium constant becomes:
\[ K'' = \sqrt{K'} = \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1 \]
Final Answer:
\[ \boxed{0.1} \]