The equilibrium constant for a reacton
`N_(2)(g)+O_(2)(g)=2NO(g)` is `4xx10^(-4)` at `2000 K`. In the presence of catalyst, the equilibrium constant is attained `10` times faster. The equilibrium constant in the presence of catalyst, at `2000 K` is

Options

A. `40xx10^(-4)`
B. `4xx10^(-4)`
C. `4xx10^(-3)`
D. difficult to compute without more data

Answer

To solve the problem, we need to understand the effect of a catalyst on the equilibrium constant of a reaction. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
We are given the equilibrium constant (K) for the reaction:
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]
at a temperature of 2000 K, which is:
\[ K = 4 \times 10^{-4} \]

### Step 2: Analyze the Effect of a Catalyst
The problem states that in the presence of a catalyst, the equilibrium is attained 10 times faster. However, it is important to note that a catalyst does not change the equilibrium constant of a reaction. It only speeds up the rate at which equilibrium is reached for both the forward and reverse reactions equally.

### Step 3: Conclusion on the Equilibrium Constant
Since the catalyst does not alter the equilibrium constant, the value of K remains the same regardless of the presence of the catalyst. Therefore, the equilibrium constant in the presence of the catalyst at 2000 K is still:
\[ K = 4 \times 10^{-4} \]

### Final Answer
The equilibrium constant in the presence of the catalyst at 2000 K is:
\[ 4 \times 10^{-4} \]

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