Problem 73 - Chemistry The Molecular Nature of Matter

Question

The reaction \(2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)\) has \(K_{\mathrm{c}}=\) \(3.2 \times 10^{-34}\) at \(25^{\circ} \mathrm{C}\). If a reaction vessel contains initially \(0.0500 \mathrm{~mol} \mathrm{~L}^{-1}\) of \(\mathrm{HCl}\) and then reacts to reach equilibrium, what will be the concentrations of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) ?

Step-by-Step Solution

Answer

The equilibrium concentrations of H2 and Cl2 are both approximately 2.83 x 10^-18 M, and the final concentration of HCl is approximately 0.0500 M.

Step 1: Interpret the Chemical Equation

First, understand the balanced chemical equation provided. Here, 2 moles of HCl gas decompose into 1 mole of H2 gas and 1 mole of Cl2 gas in a reversible reaction. The reaction quotient, or Kc, represents the equilibrium concentrations of the products raised to their stoichiometric coefficients, divided by the concentrations of the reactants raised to their stoichiometric coefficients.

Step 2: Write the Expression for Kc

The equilibrium constant expression (Kc) for the reaction is written as: \[ K_c = \frac{{[\mathrm{H}_2][\mathrm{Cl}_2]}}{{[\mathrm{HCl}]^2}} \]

Step 3: Set up an ICE Table

Create an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations. The initial concentration of HCl is 0.0500 M, and H2 and Cl2 are initially 0. The change for HCl is -2x (since 2 moles of HCl decompose), and +x for both H2 and Cl2. At equilibrium, the concentrations are [HCl] = 0.0500 - 2x, [H2] = x, and [Cl2] = x.

Step 4: Substitute the Equilibrium Concentrations into the Kc Expression

Put the expressions from the ICE table into the Kc expression, \[ K_c = 3.2 \times 10^{-34} = \frac{{x^2}}{{(0.0500 - 2x)^2}} \]. Because Kc is very small, we can assume that x will be much smaller than 0.0500, which allows us to simplify the denominator to 0.0500^2.

Step 5: Solve for x

After the simplification, the equation looks like \[ 3.2 \times 10^{-34} = \frac{{x^2}}{{(0.0500)^2}} \]. Solve for x to find the equilibrium concentration of H2 and Cl2.

Step 6: Calculate the Equilibrium Concentrations

The concentration of H2 and Cl2 at equilibrium are both x. \[ x^2 = 3.2 \times 10^{-34} \times (0.0500)^2 \], \[x = \sqrt{3.2 \times 10^{-34} \times (0.0500)^2}\], \[ x = \sqrt{8 \times 10^{-36}} \], \[ x = 2.83 \times 10^{-18} M \]. This is the equilibrium concentration of both H2 and Cl2.

Step 7: Confirm the Approximation

Confirm that the assumption x << 0.0500 is valid by checking if 2x is much less than the initial concentration of HCl. If the assumption is valid, then the calculated x is the correct equilibrium concentration.

Step 8: Calculate the Final Concentration of HCl

Subtract 2x from the initial concentration of HCl to get its equilibrium concentration. \[ [\mathrm{HCl}]_{eq} = 0.0500 M - 2(2.83 \times 10^{-18} M) \], \[ [\mathrm{HCl}]_{eq} = 0.0500 M - 5.66 \times 10^{-18} M \approx 0.0500 M \]. This is negligibly different from the initial concentration, which confirms our approximation was reasonable.

Key Concepts

Equilibrium Constant Expression

In chemistry, understanding the equilibrium constant expression is crucial when dealing with reversible reactions. This expression, commonly denoted as Kc, relates to the concentrations of products and reactants at chemical equilibrium. It is given by the ratio of the product of equilibrium concentrations of products, each raised to the power of its coefficient in the balanced equation, to the product of equilibrium concentrations of reactants similarly raised to the power of their coefficients. For the reaction \(2 \text{HCl}_{(g)} \rightleftharpoons \text{H}_2_{(g)} + \text{Cl}_2_{(g)}\),the equilibrium constant expression is \[ K_c = \frac{{[\text{H}_2][\text{Cl}_2]}}{{[\text{HCl}]^2}} \].In this reaction, notice the squared term for HCl due to its stoichiometric coefficient of 2. The equilibrium constant, Kc, has a value specific to a particular reaction at a given temperature, which in this case is \(3.2 \times 10^{-34}\) at \(25^\circ C\).

ICE Table Method

The ICE table method is an invaluable tool that stands for Initial, Change, and Equilibrium. It's used to organize and calculate the changes in concentration that occur when a system reaches equilibrium. To employ this method, you start by listing initial concentrations, followed by the changes that occur as reactants convert to products, and then the final equilibrium concentrations. For the reaction in question, the initial concentration of HCl is given, while those of H2 and Cl2 are zero.

The next step involves representing the changes in concentration using variables, often 'x'. The stoichiometry of the balanced equation informs the relationship between the changes for each species. In this case, since 2 moles of HCl convert into 1 mole each of H2 and Cl2, we represent the decrease in HCl as -2x and the increase for H2 and Cl2 as +x. When equilibrium is reached, the concentrations are adjusted by these changes, resulting in the final equilibrium concentrations. This systematic approach simplifies understanding the progression from initial to equilibrium conditions.

Reaction Quotient Kc

The reaction quotient, denoted as Qc, is similar to the equilibrium constant, Kc, but for a system that has not yet reached equilibrium. It uses the same formula as Kc but with the current concentrations instead of the equilibrium concentrations. It's a snapshot of a reaction's status. If Qc equals Kc, the system is at equilibrium. If Qc is greater than Kc, the system will shift toward the reactants to reach equilibrium, and if Qc is less than Kc, the reaction will proceed toward the products to achieve equilibrium. By calculating Qc at any point during a reaction, you can predict the direction the reaction will shift to reach equilibrium, providing valuable insight into the reaction's behavior.

Stoichiometry

The concept of stoichiometry is rooted in the balanced chemical equation and informs all calculations concerning chemical reactions. It's the arithmetic relationship between reactants and products, defined by their coefficients in a balanced equation. This relationship tells us, for example, that for every two moles of HCl that decompose, one mole each of H2 and Cl2 are formed. Correct stoichiometric calculation is critical when setting up the ICE table, as it ensures accurate representation of the concentration changes as the reaction progresses toward equilibrium. Being adept in stoichiometry allows you to confidently determine how reactants will convert into products and vice versa, a fundamental skill in chemistry.

Equilibrium Concentration Calculation

Calculating equilibrium concentrations is the final step in understanding chemical equilibrium. Using the ICE table and the equilibrium constant expression, we derive a mathematical relationship we can use to solve for the unknown concentration(s). Once the equation is set up with the expression for Kc and the known initial concentrations, we can make reasonable approximations to simplify the math if Kc is very small or very large, letting us focus on the major changes. This process involves solving for 'x', which represents the change in concentration. The accuracy of the result hinges on correct stoichiometric coefficients and the justified approximations made during the calculation. The calculated equilibrium concentrations provide not only a quantitative understanding of the reaction at equilibrium but also validate the assumptions made, like changes being negligible compared to initial concentrations. This enables chemists to predict the behavior of reactions under various initial conditions and to optimize them accordingly.