Question

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

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

Write the equilibrium expression

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

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: \(K = \frac{(4.7 \times 10^{-4})^2}{(0.041)(0.0078)}\)

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 \(9.13 \times 10^{-4}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry that explains how reactions reach a state of balance. In a chemical reaction, reactants convert into products over time. But, unlike a race to an endpoint, these reactions can reach a point where the rate of the forward reaction (making products) equals the rate of the reverse reaction (making reactants). This state is called chemical equilibrium.
At equilibrium, the concentrations of reactants and products remain constant. They are not equal, but they stabilize at a specific ratio. It's like two equally strong teams playing tug-of-war, ending in a steady hold where neither team moves.
This balance, however, doesn't mean the reactions stop. They continue to occur, but since they occur at the same rate, there is no net change in concentrations. Understanding this principle is key to calculating the equilibrium constant and understanding reaction dynamics.

Reaction Quotient

The reaction quotient, often represented by the letter 'Q,' is a valuable tool when analyzing chemical reactions. It's similar to the equilibrium constant (K) but it doesn't require the system to be at equilibrium.
The same formula applies to both Q and K: \[ Q = \frac{[\text{products}]^x}{[\text{reactants}]^y} \] where \(x\) and \(y\) are the stoichiometric coefficients from the balanced chemical equation.
By comparing Q and K, we can predict the direction in which a reaction will proceed to reach equilibrium:

• If \(Q < K\), the reaction will proceed forward to produce more products.
• If \(Q > K\), the reaction will shift backward to form more reactants.
• If \(Q = K\), the system is at equilibrium.

Understanding the reaction quotient is useful for foreseeing how a reaction needs to be adjusted to reach equilibrium. In our example, since we had the equilibrium concentrations, we directly calculated \(K\) to confirm the system's state.

Concentration Calculations

Concentration calculations are straightforward once you know how to use the formulae correctly. Performing these calculations allows us to understand how much of each species is present in a reaction mixture. For equilibrium problems, this involves using the equilibrium expression.
In our original problem, we were given equilibrium concentrations:

• \([\text{N}_2] = 0.041 \, M\)
• \([\text{O}_2] = 0.0078 \, M\)
• \([\text{NO}] = 4.7 \times 10^{-4} \, M\)

We plugged these values into the equilibrium expression \( K = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \).This calculation determines the equilibrium constant \(K\), giving us a specific number (\(9.13 \times 10^{-4}\)) that describes how far the reaction has proceeded under the given conditions.
These calculations are critical for predicting the behavior of chemical reactions and for industrial applications where maintaining specific concentration ratios is necessary.