Question

Consider the following reaction and its equilibrium constant at \(100^{\circ} \mathrm{C}\) : $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad K_{\mathrm{eq}}=6.5 $$ If \(0.250 \mathrm{~mol}\) of each reactant and product is mixed into a \(1.0\) - \(\mathrm{L}\) container, will the reaction proceed in the forward or reverse direction, or is it already at equilibrium?

Short answer

The reaction will proceed in the forward direction.

Step-by-step solution

Write Down the Reaction

Start with writing down the given chemical reaction: \[ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \] and it is also given that at equilibrium \(K_{\mathrm{eq}}=6.5\)

Calculate the Reaction Quotient Q

The reaction quotient is calculated in the same manner as the equilibrium constant but uses initial concentrations instead of equilibrium concentrations. For the given reaction, the formula for the reaction quotient Q is \(Q= \frac {[NO_{2}]^2} {[N_{2}O_{4}]}\). Now, substitute the given initial concentrations into the Q formula (with \([N_{2}O_{4}]= [NO_{2}]= 0.250 M\), as initially, 0.250 mol of each were mixed into a 1.0-L container): \[ Q= \frac {(0.250)^2}{0.250} = 0.250\]

Compare Q and K

Now, compare the calculated Q value with the given K value. If Q < K, the reaction will proceed in the forward direction to reach equilibrium; if Q > K, the reaction will go in the reverse direction; if Q = K, the reaction is already at equilibrium. In this case, since \[0.250 < 6.5\], the reaction will proceed in the forward direction. This means more of the reactant will turn into the products until equilibrium is reached.

Key concepts

Understanding the Reaction Quotient (Q)

The reaction quotient, commonly denoted as \( Q \), is a crucial concept in chemical equilibrium. It helps us to understand the state of a chemical reaction at any given point, not necessarily at equilibrium. This is particularly useful in determining the direction a reaction will shift to achieve equilibrium.

To calculate \( Q \), we use the same mathematical expression as for calculating the equilibrium constant \( K_{eq} \). The key difference is that \( Q \) uses the initial concentrations of the reactants and products rather than the concentrations at equilibrium.

For example, the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \) uses the formula for \( Q \) as \( Q = \frac{[NO_{2}]^2}{[N_{2}O_{4}]} \). Here, substituting the initial concentrations as given (both are \( 0.250 \text{ M} \)), the calculated \( Q \) is \( 0.250 \).

By comparing this with \( K_{eq} \), we can predict the movement of the reaction to achieve equilibrium.

Equilibrium Constant (K_eq) Explained

The equilibrium constant \( K_{eq} \) is a significant factor in understanding chemical reactions at a state of balance. It is essentially a measure of the ratio of products to reactants when a reaction is at equilibrium.

Each reaction has a unique \( K_{eq} \), and it is derived from the concentrations of the reactants and products once the reaction has reached equilibrium at a given temperature.

In our example reaction, \( N_{2}O_{4} \rightleftharpoons 2 NO_{2} \), the \( K_{eq} \) is given as \( 6.5 \) at \( 100^{\circ} \mathrm{C} \). This means that at equilibrium, the products \( NO_{2} \) are more favored over the reactants \( N_{2}O_{4} \) given that there is a higher ratio of products to reactants.

A higher \( K_{eq} \) indicates a greater concentration of products at equilibrium, whereas a lower \( K_{eq} \) suggests a favor towards reactants. This helps to predict which direction the reaction will shift based on the initial concentrations.

Factors Influencing Forward and Reverse Reactions

Chemical reactions are dynamic, meaning they can proceed in both forward and reverse directions. Initially, whether a reaction will proceed in one direction more than the other greatly depends on the relative magnitudes of \( Q \) and \( K_{eq} \).

When \( Q < K_{eq} \), the reaction progresses in the forward direction. This means more reactants are converted to products, as seen in our example with \( 0.250 < 6.5 \), requiring more \( N_{2}O_{4} \) to turn into \( 2 NO_{2} \) to reach equilibrium.

When \( Q > K_{eq} \), the reaction moves in the opposite reverse direction, converting more products back into reactants. Lastly, if \( Q = K_{eq} \), the system is already at equilibrium and there will be no net change in the concentrations of reactants and products.

Understanding these concepts is vital to predict the outcome of reactions and to manipulate conditions in practical applications to favor the desired products.