Problem 55 - Chemistry

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 \(\left[\mathrm{H}_{2}\right]=5.0 M,\left[\mathrm{N}_{2}\right]=\) \(8.0 M,\) and \(\left[\mathrm{NH}_{3}\right]=4.0 \mathrm{M} .\) What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

Step-by-Step Solution

Answer

The initial concentration of hydrogen gas (H2) was 11 mol/L, and the initial concentration of nitrogen gas (N2) was 10 mol/L.

Step 1: Set up the equilibrium constant and balance the equation

First, balance the given chemical equation: \[ 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \] Then, write down the equilibrium constant expression for this reaction: \[ K_{c} = \frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{H}_{2}\right]^{3} \left[\mathrm{N}_{2}\right]} \]

Step 2: Let the initial concentrations be x and y

Let the initial concentrations of hydrogen gas and nitrogen gas be x mol/L and y mol/L, respectively.

Step 3: Set up the reaction table for the stoichiometry

Set up an ICE (Initial, Change, Equilibrium) table to represent the changes from the initial concentrations to the equilibrium concentrations: | Substance | Initial (mol/L) | Change (mol/L) | Equilibrium (mol/L) | |---|---|---|---| | H2 | x | -3z | 5 | | N2 | y | -z | 8 | | NH3 | 0 | 2z | 4 | Where z represents the change that occurred in the concentrations during the reaction.

Step 4: Calculate z from the given equilibrium concentrations

We can now use the equilibrium concentrations to solve for z: \[5 = x - 3z\] \[8 = y - z\] \[4 = 2z\] From the last equation, we can find the value of z: \[ z = \frac{4}{2} = 2 \]

Step 5: Find the initial concentrations of H2 and N2

Using the value of z, we can now find the initial concentrations of hydrogen gas (x) and nitrogen gas (y): For H2: \[ x = 5 + 3z = 5 + 3(2) = 5 + 6 = 11 \ \text{mol/L} \] For N2: \[ y = 8 + z = 8 + 2 = 10 \ \text{mol/L} \] So, the initial concentration of hydrogen gas was 11 mol/L, and the initial concentration of nitrogen gas was 10 mol/L.

Key Concepts

Equilibrium Constant

In chemical reactions that reach a state of balance, the equilibrium constant (K_c) helps us determine the position of equilibrium. This constant is derived from the law of mass action, and it's a ratio that compares the concentrations of products to reactants at equilibrium. For the reaction provided:

• The balanced equation is \(3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\).
• We form the equilibrium constant expression as \(K_{c} = \frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{H}_{2}\right]^{3} \left[\mathrm{N}_{2}\right]}\).

This powerful expression allows scientists to understand how a mixture of substances behaves over time.

• If \(K_c\) is much greater than 1: The equilibrium lies toward the products.
• If \(K_c\) is much smaller than 1: The equilibrium favors the reactants.
• An equilibrium constant close to 1 suggests a significant amount of both reactants and products.

ICE Table

The ICE table is a systematic way to visualize how the concentrations of reactants and products change during a reaction. ICE stands for Initial, Change, and Equilibrium, and it's set out in a tabular form:

• Initial line: shows the starting concentrations of all species.
• Change line: indicates how much each substance increases or decreases.
• Equilibrium line: displays the final concentrations once equilibrium is achieved.

For the exercise example, the table looks like this:- **H2** initially at \(x\), changes by \(-3z\), ending at 5 M.- **N2** initially at \(y\), changes by \(-z\), ending at 8 M.- **NH3** starts at 0, increases by \(+2z\), and ends at 4 M.This table allows us to predict and calculate the equilibrium concentrations and the changes involved, assisting in determining initial concentrations of the reactants.

Initial Concentrations

The initial concentrations of substances tell us how much of each reactant or product is present before the reaction starts. In equilibrium problems, knowing these values is crucial for setting up the reaction's progress:- For our problem, the variables \(x\) and \(y\) represent the initial concentrations of **H2** and **N2**, respectively.- By using the ICE table's equilibrium data, we work backward to uncover these starting values.To find these, we need to plug in the value of \(z\) (which indicates the change in concentrations during the reaction) back into the initial expressions:

• For **H2**: \(x = 5 + 3z\)
• For **N2**: \(y = 8 + z\)

Here, once \(z\) was found to be 2, it was simple to calculate:- **H2** started at 11 M.- **N2** began at 10 M.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry allows us to determine the proportional relationships between reactants and products in a reaction. This is guided by the balanced chemical equation, ensuring _matter conservation_ throughout:The balanced equation \(3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) uses stoichiometric coefficients to guide the changes:

• This tells us for every 1 mole **N2** consumed, 3 moles **H2** are consumed as well.
• Similarly, 2 moles of **NH3** are formed.

Stoichiometry was fundamental in setting up the ICE table where we represented changes with \(z\):- \(-3z\) for H2 to reflect the three-to-one ratio with NH3's formation.- \(+2z\) for NH3 as each unit needed 3 units of H2.By understanding stoichiometry, we accurately describe how quantities leverage each other during the reaction, paving the way for precise calculations of initial and equilibrium concentrations.