Question

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ is 54.3 at \(430^{\circ} \mathrm{C}\). At the start of the reaction there are 0.714 mole of \(\mathrm{H}_{2}, 0.984\) mole of \(\mathrm{I}_{2}\), and 0.886 mole of HI in a 2.40-L reaction chamber. Calculate the concentrations of the gases at equilibrium.

Short answer

Therefore, the concentrations of the gases at equilibrium are \( [H_2]_{eq} = 0.201 M \), \( [I_2]_{eq} = 0.313 M\) and \([HI]_{eq} = 0.563 M\).

Step-by-step solution

Write initial concentrations

First calculate the initial concentrations of the species in the equation by dividing the number of moles of each by the total volume of the reaction chamber. This gives: \[ [H_2]_{initial} = \frac{0.714 mol}{2.40 L} = 0.298 M \] \[ [I_2]_{initial} = \frac{0.984 mol}{2.40 L} = 0.410 M \] \[ [HI]_{initial} = \frac{0.886 mol}{2.40 L} = 0.369 M \]

Write the expression for \(K_{c}\)

The next step is to write the expression for the equilibrium constant \(K_{c}\). For the given reaction, this is: \[ K_{c} = \frac{[HI]_{eq}^2}{[H_2]_{eq}[I_2]_{eq}} \]

Set up a table of concentrations

We can now set up a table to calculate the equilibrium concentrations of all the species, called an ICE (Initial-Change-Equilibrium) Table. As the reaction proceeds forward, the concentrations of \(H_2\) and \(I_2\) decrease by \(x\) while that of HI increases by \(2x\). Thus, the equilibrium concentrations are: \[ [H_2]_{eq} = 0.298 - x \] \[ [I_2]_{eq} = 0.410 - x \] \[ [HI]_{eq} = 0.369 + 2x \]

Solve for \(x\)

Plugging these into the \(K_{c}\) expression, we have: \[ 54.3 = \frac{(0.369+2x)^2}{(0.298-x)(0.410-x)} \] This equation can be solved to find the value of \(x\), which comes out to be 0.097.

Calculate equilibrium concentrations

Finally substituting the value of \(x\), the equilibrium concentrations can be found: \[ [H_2]_{eq} = 0.298 - 0.097 = 0.201 M \] \[ [I_2]_{eq} = 0.410 - 0.097 = 0.313 M\] \[ [HI]_{eq} = 0.369 + 2*(0.097) = 0.563 M \]

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in chemical reactions where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. When a reaction reaches this state, it may appear to have stopped, but in reality, both reactions are still occurring, just at the same rate. This dynamic balance allows the concentrations of all reactants and products to remain constant at equilibrium.

For the reaction \(H_{2}(g) + I_{2}(g) \rightleftharpoons 2 HI(g)\), equilibrium will be achieved when the rate at which \(H_{2}\) and \(I_{2}\) react to form \(HI\) is equal to the rate at which \(HI\) decomposes to form \(H_{2}\) and \(I_{2}\). The state of equilibrium can be disturbed by changing concentration, pressure, or temperature, leading to a shift in position as predicted by Le Châtelier's principle.

ICE Table Method

The ICE Table method is an organized approach for calculating the changes in concentrations of species in a reaction as they move towards equilibrium. ICE stands for Initial, Change, Equilibrium. It helps in visualizing how the concentrations of reactants and products change from their initial values (I) to the new values at equilibrium (E) as some of the reactants are converted into products (Change, C).

Using the ICE Table, we can systematically calculate equilibrium concentrations by setting up a table with rows for each substance and columns for initial concentrations, changes in concentrations, and equilibrium concentrations. This method serves as a visual aid to keep track of the stoichiometry of the reaction and the changes that occur as the reaction moves towards equilibrium. For reactions where the stoichiometry is one-to-one, this method is typically straightforward. However, it becomes particularly useful for reactions with more complex stoichiometries.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of a reactant or product in a reaction mixture when the reaction has reached equilibrium. At this point, the concentrations are stable and do not change over time. In the exercise, after setting up the ICE table, we found that equilibrium concentrations for \(H_{2}\), \(I_{2}\), and HI were \(0.201 M\), \(0.313 M\), and \(0.563 M\) respectively. These concentrations are calculated by taking the initial concentrations of each species and adjusting them by the amount they have changed due to the reaction proceeding towards equilibrium (denoted by \(x\) in the ICE table).

Importance in Reactions

Understanding equilibrium concentrations is crucial for chemists. These values can be used to determine reaction yields, optimize reaction conditions, and calculate the equilibrium constant of a reaction under specific temperatures or pressures.

Reaction Quotient

The reaction quotient (Q) is a measure that tells us the direction in which a reaction mixture will shift to reach equilibrium. It is calculated using the same formula as the equilibrium constant (\(K\)), but with the initial concentrations instead of the equilibrium concentrations. If \(Q < K\), the reaction will proceed forward to produce more products. If \(Q > K\), the reaction will shift in reverse to produce more reactants. When \(Q = K\), the reaction is at equilibrium and no further shift will occur.

For the given exercise, once we have the initial concentrations, we can calculate \(Q\) to predict the direction of the shift. However, since the equilibrium constant (\(K_{c}\)) is provided, and we need to determine the equilibrium concentrations for a reaction that has already reached equilibrium, the reaction quotient is not calculated directly in this step but is implied to be equal to \(K\). The ultimate goal is to find the value of \(x\) that brings the reaction quotient in line with the given equilibrium constant.