Question

At \(350^{\circ} \mathrm{C}, K_{\mathrm{eq}}=1.67 \times 10^{-2}\) for the reversible reaction \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}^{2}(\mathrm{g})+\mathrm{I}^{2}(\mathrm{g}) .\) What is the concentration of HI at equilibrium if \(\left[\mathrm{H}^{2}\right]\) is \(2.44 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{I}^{2}\right]\) is \(7.18 \times 10^{-5} \mathrm{M} ?\)

Short answer

The concentration of HI at equilibrium can be calculated using the given equilibrium constant, \(K_{eq}\), and the concentrations of H2 and I2 at equilibrium. By plugging these values into the equilibrium constant expression, \([\mathrm{HI}] = \sqrt{\dfrac{(2.44 \times 10^{-3})(7.18 \times 10^{-5})}{1.67 \times 10^{-2}}}\). The concentration of HI at equilibrium is approximately 0.00494 M.

Step-by-step solution

Write the equilibrium constant expression

The expression for the equilibrium constant, \(K_{eq}\), for the given reversible reaction can be found using the concentrations of the products over the reactants. For the reaction: \(2\mathrm{HI(g)} \rightleftharpoons \mathrm{H}^{2}\mathrm{(g)}+\mathrm{I}^{2}\mathrm{(g)}\) The equilibrium constant expression is: \[K_{eq} = \dfrac{[\mathrm{H}^2][\mathrm{I}^2]}{[\mathrm{HI}]^2}\]

Plug in the given concentrations and equilibrium constant

We are given the values of \(K_{eq} = 1.67 \times 10^{-2}\), \([\mathrm{H}^{2}] = 2.44 \times 10^{-3} \mathrm{M}\), and \([\mathrm{I}^{2}] = 7.18 \times 10^{-5} \mathrm{M}\). Plug these values into the equilibrium constant expression: \(1.67 \times 10^{-2} = \dfrac{(2.44 \times 10^{-3})(7.18 \times 10^{-5})}{[\mathrm{HI}]^2}\)

Solve for the concentration of HI

Rearrange the equation to isolate \([\mathrm{HI}]\) on one side and then solve for the concentration of HI: \[\mathrm{[HI]}^2 = \dfrac{(2.44 \times 10^{-3})(7.18 \times 10^{-5})}{1.67 \times 10^{-2}}\] Now, take the square root of both sides: \[\mathrm{[HI]} = \sqrt{\dfrac{(2.44 \times 10^{-3})(7.18 \times 10^{-5})}{1.67 \times 10^{-2}}}\] \[\mathrm{[HI]} \approx 0.00494 \mathrm{M}\] So, the concentration of HI at equilibrium is approximately 0.00494 M.

Key concepts

Equilibrium Constant

The equilibrium constant (\( K_{eq} \)) is a crucial concept in understanding chemical equilibrium. It quantifies the ratio of product concentrations to reactant concentrations at equilibrium. This value helps us predict the direction of the reaction and how the concentrations will shift.
For the reaction \( 2\mathrm{HI(g)} \rightleftharpoons \mathrm{H}^{2}\mathrm{(g)}+\mathrm{I}^{2}\mathrm{(g)} \), \( K_{eq} \) is given by the expression:
\[K_{eq} = \dfrac{[\mathrm{H}^2][\mathrm{I}^2]}{[\mathrm{HI}]^2}\]
This formula indicates:

• The concentrations of the products \( [\mathrm{H}^2] \) and \( [\mathrm{I}^2] \), are in the numerator.
• The concentration of the reactant \( [\mathrm{HI}] \), raised to the power of its coefficient (2 in this case), is in the denominator.

The equilibrium constant is temperature-dependent, meaning \( K_{eq} \) can change with different temperatures. For the reaction at \( 350^{\circ} \mathrm{C} \), \( K_{eq} \) is \( 1.67 \times 10^{-2} \). This relatively small value indicates a preference towards the reactants' side, where concentrations of products are not overwhelmingly high compared to the reactants.

Reversible Reactions

Reversible reactions are central to the concept of chemical equilibrium. A reversible reaction means that the reaction can proceed in both the forward and reverse directions. Over time, these reactions can reach a state of dynamic equilibrium where the rate of the forward reaction equals the rate of the backward reaction. This equilibrium state does not imply that the concentrations of reactants and products are equal, but rather that their rates are balanced.
Consider the reaction \( 2\mathrm{HI(g)} \rightleftharpoons \mathrm{H}^{2}\mathrm{(g)}+\mathrm{I}^{2}\mathrm{(g)} \).

• In the forward reaction, \( 2\mathrm{HI} \) molecules produce \( \mathrm{H}^2 \) and \( \mathrm{I}^2 \).
• In the reverse reaction, \( \mathrm{H}^2 \) and \( \mathrm{I}^2 \) recombine to form \( 2\mathrm{HI} \).

Achieving equilibrium means that despite ongoing reactions in both directions, the system's overall energy state remains constant. This allows us to use the equilibrium constant to calculate unknown concentrations or to predict changes in the system when disturbed.

Concentration Calculations

Concentration calculations at equilibrium give us the values of reactants or products when a system has reached equilibrium. Using the equilibrium constant expression, we can solve for unknown concentrations.
In our reaction at \( 350^{\circ} \mathrm{C} \), we were given \( K_{eq} = 1.67 \times 10^{-2} \), with\( [\mathrm{H}^2] = 2.44 \times 10^{-3} \mathrm{M} \) and \( [\mathrm{I}^2] = 7.18 \times 10^{-5} \mathrm{M} \).
To find the concentration of \( [\mathrm{HI}] \), we start with the equilibrium constant expression:
\[1.67 \times 10^{-2} = \dfrac{(2.44 \times 10^{-3})(7.18 \times 10^{-5})}{[\mathrm{HI}]^2}\]Rearranging this equation to solve for \( [\mathrm{HI}] \), we find:
\[[\mathrm{HI}]^2 = \dfrac{(2.44 \times 10^{-3})(7.18 \times 10^{-5})}{1.67 \times 10^{-2}}\]After solving for \( [\mathrm{HI}] \), we take the square root:\[[\mathrm{HI}] = \sqrt{\dfrac{(2.44 \times 10^{-3})(7.18 \times 10^{-5})}{1.67 \times 10^{-2}}}\]This results in \([\mathrm{HI}] \approx 0.00494 \mathrm{M}\).
These calculations show how to utilize given data and the equilibrium constant to discover unknown concentrations in a chemical reaction at equilibrium.