At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 M,\left[\mathrm{O}_{2}\right]=0.0078 M,\) and $[\mathrm{NO}]=4.7 \times 10^{-4} M .\( Calculate the value of \)K$ for the reaction.

Short Answer

Expert verified
The equilibrium constant (K) for the reaction \(\mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) can be calculated using the given equilibrium concentrations: \([\mathrm{N}_{2}] = 0.041 M\), \([\mathrm{O}_{2}] = 0.0078 M\), and \([\mathrm{NO}] = 4.7 \times 10^{-4} M\). By plugging these values into the equilibrium expression \(K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]}\), we find that K is approximately \(9.13 \times 10^{-4}\).

Step by step solution

01

Write the equilibrium expression

The equilibrium expression for the given reaction is: \(K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]}\)
02

Plug in the given equilibrium concentrations

We are given the equilibrium concentrations as follows: \([\mathrm{N}_{2}] = 0.041 M\) \([\mathrm{O}_{2}] = 0.0078 M\) \([\mathrm{NO}] = 4.7 \times 10^{-4} M\) Now, we plug these values into the equilibrium expression: \(K = \frac{(4.7 \times 10^{-4})^2}{(0.041)(0.0078)}\)
03

Calculate K

Now, let's calculate the value of K: \(K = \frac{(4.7 \times 10^{-4})^2}{(0.041)(0.0078)} \approx 9.13 \times 10^{-4}\) Thus, the equilibrium constant (K) for the given reaction is approximately \(9.13 \times 10^{-4}\).

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