# Calculate the equilibrium concentration of "NO" if the equilibrium concentration of "N"_2 is "0.036mol L"^(-1) and that of "O"_2 is "0.0089 mol L"^(-1) ?

## What is the equilibrium constant at ${2000}^{\circ} \text{C}$ for the reaction $\text{N"_2(g) + "O"_2(g)-> 2"NO} \left(g\right)$ is $4.1 \cdot {10}^{4}$.

##### 1 Answer
Feb 12, 2018

${\text{3.6 mol L}}^{- 1}$

#### Explanation:

You know that at ${2000}^{\circ} \text{C}$, the reaction

${\text{N"_ (2(g)) + "O"_ (2(g)) -> 2"NO}}_{\left(g\right)}$

has an equilibrium constant equal to

${K}_{c} = 4.1 \cdot {10}^{4}$

Now, the equilibrium constant is calculated by taking the equilibrium concentrations of the chemical species involved in the reaction.

${K}_{c} = \left(\left[{\text{NO"]^2)/(["N"_2] * ["O}}_{2}\right]\right)$

At equilibrium, you have

• ["N"_2] = "0.036 mol L"^(-1)
• ["O"_2] = "0.0089 mol L"^(-1)

Rearrange the equation to isolate the equilibrium concentration of nitric oxide.

["NO"] = sqrt( K_c * ["N"_2] * ["O"_2])

Plug in your values to find

$\left[{\text{NO"] = sqrt(4.1 * 10^4 * "0.036 mol L"^(-1) * "0.0089 mol L}}^{- 1}\right)$

color(darkgreen)(ul(color(black)(["NO"] = "3.6 mol L"^(-1))))

The answer is rounded to two sig figs.

Notice that the equilibrium concentration of nitric oxide is significantly higher than the equilibrium concentrations of the nitrogen gas and the oxygen gas.

This is the case because you have

${K}_{c} \text{>>} 1$

which implies that the equilibrium lies to the right, i.e. at this temperature, the forward reaction is favored.