Question

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction $$ 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ At equilibrium, the concentrations are \(\left[\mathrm{H}_{2}\right]=5.0 \mathrm{M},\left[\mathrm{N}_{2}\right]=\) \(8.0 M\), and \(\left[\mathrm{NH}_{3}\right]=4.0 M .\) What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

Short answer

The initial concentrations of nitrogen gas and hydrogen gas were 9 M and 6 M, respectively.

Step-by-step solution

The balanced chemical equation for this reaction is: \( 3 H_2(g) + N_2(g) \rightleftharpoons 2 NH_3(g) \) #Step 2: Write the expression for the equilibrium constant (K) #

The expression for the equilibrium constant (K) can be written as: \( K = \frac{[NH_3]^2}{([H_2]^3 [N_2])} \) #Step 3: Calculate K using the given equilibrium concentrations#

We are given the equilibrium concentrations: [ \( H_2 \) ] = 5.0 M [ \( N_2 \) ] = 8.0 M [ \( NH_3 \) ] = 4.0 M Then, calculate K: \( K = \frac{4.0^2}{(5.0^3 \times 8)} \) \( K = \frac{16}{(125 \times 8)} \) \( K = \frac{16}{1000} = 0.016 \) #Step 4: Set up an ICE table (Initial, Change, Equilibrium)#

Let x be the change in the concentration of reactants and products from their initial concentrations to at the equilibrium. [ \( H_2 \) ] [ \( N_2 \) ] [ \( NH_3 \) ] Initial: 5+x 8+x 0-2x Change: -3x -x +2x Equilibrium: 5 8 4 #Step 5: Substitute the concentrations back into the K expression#

Substitute the equilibrium concentrations back into the K expression and solve for x: \( K = 0.016 = \frac{(4)^2}{((5+x)^3 \times (8+x))} \) #Step 6: Solve for x and find initial concentrations#

As this equation can be hard to solve algebraically, we will approximate the solution (assuming x is small compared to 5 and 8): \( 0.016 = \frac{16}{(5^3 \times 8)} \) Solve for x: \( x = \frac{16}{(125 \times 8 \times 0.016 )} \) \( x \approx 1 \) Initial concentrations: [ \( H_2 \) ] = 5 + x = 5 + 1 = 6 M [ \( N_2 \) ] = 8 + x = 8 + 1 = 9 M Hence, the initial concentrations of nitrogen gas and hydrogen gas were 9 M and 6 M, respectively.

Key concepts

Chemical Equilibrium

Chemical equilibrium is the state of a reversible chemical reaction where the rate of the forward reaction is equal to the rate of the backward reaction. In this state, the concentrations of reactants and products stop changing with time.
It's essential to understand that at equilibrium, both reactants and products coexist in the reaction mixture. However, their concentrations remain constant and do not necessarily mean they are equal.
The equilibrium constant, denoted as \( K \), is a measure of the extent to which a reaction proceeds to form products at equilibrium. A large \( K \) value implies more products at equilibrium, whereas a small \( K \) value indicates more reactants.
In the exercise, nitrogen and hydrogen react to form ammonia (\( NH_3 \)), and at equilibrium, specific concentrations are reached. These concentrations are used to compute the equilibrium constant \( K \), which helps us understand the ratio of products to reactants at equilibrium.

ICE Table

An ICE table is a useful method to organize and visualize changes in concentrations of reactants and products during a reaction that reaches equilibrium. "ICE" stands for Initial, Change, and Equilibrium.
These tables help break down the reaction into its initial state, how much it changes, and its equilibrium state. By laying out these values, actors in a chemical reaction can be better understood and calculated.
In the given problem, an ICE table is used to track the concentration changes of \( H_2 \), \( N_2 \), and \( NH_3 \) from their initial states to equilibrium. It clearly shows the changes, governed by stoichiometry, from initial to equilibrium concentrations.
The table structure helps in setting an equation for the equilibrium constant, incorporating the changes in concentrations as represented by \( x \) in the reaction.

Reaction Stoichiometry

Reaction stoichiometry involves using coefficients from the balanced chemical equation to relate the amounts of reactants and products. It dictates how each component of the reaction changes when reaching equilibrium.
In a balanced reaction equation, like \( 3 \mathrm{H}_2 + \mathrm{N}_2 \rightleftharpoons 2 \mathrm{NH}_3 \), the coefficients (3, 1, and 2) show the mole ratios in which these substances react and form.
These ratios are critical when using the ICE table, as they dictate how the concentration of each component changes. For instance, a decrease in \( H_2 \) would be three times that of \( NH_3 \)'s increase.
Understanding stoichiometry is key to accurately calculating the changes in concentration necessary for predicting or interpreting reaction behavior, particularly when quantifying equilibrium shifts.

Initial Concentrations

Initial concentrations refer to the amounts of reactants and/or products present before any reaction has started. Knowing these values is vital for calculating the changes in concentration as the system reaches equilibrium.
They are especially important in setting up the ICE table since they provide the starting point for tracking changes. By identifying initial values, and knowing the equilibrium concentrations, we can calculate how much reactants and products shift during the reaction.
In the problem, by approximating the change \( x \), we can reverse-calculate these initial concentrations. The calculated \( x \) shows the extent of reaction progress, helping establish how much of \( N_2 \) and \( H_2 \) were present initially before reaching the equilibrium state.
Initial concentrations are like the starting point in an equation that, combined with the stoichiometric relationships, drive the completion of equilibrium calculations.