Problem 90 - Chemistry

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.

Step-by-Step Solution

Answer

Solving the equation in Step 4 will lead to the value of x. This represents the change in concentration from the initial conditions to equilibrium state of all the gases. Substituting x in the equation from step 3 will give the equilibrium concentrations of \(H2(g)\), \(I2(g)\), and \(HI(g)\).

Step 1: Understand the Balance Chemical Equation

The given balanced chemical equation is \(H2(g) + I2(g) \rightleftharpoons 2 HI(g)\). It indicates that 1 mole of \(H2(g)\) reacts with 1 mole of \(I2(g)\) to produce 2 moles of \(HI(g)\). Since the given reaction reaches equilibrium, this means that the concentrations of reactants and products become constant over time.

Step 2: Calculate Initial Concentrations of each Gas

The initial concentration of each gas can be calculated using the formula for molarity, which is \(M=\frac{Number\:of\:Moles}{Volume}\). Given the initial moles and reaction chamber volume, these are: \[ [H2]_i = \frac{0.714\: mol}{2.4\: L} = 0.2975 \: M , [I2]_i = \frac{0.984\: mol}{2.4 \:L} = 0.410 \:M , [HI]_i = \frac{0.886 \: mol}{2.4 \: L} = 0.369 \: M \]

Step 3: Set up the Confirmation for Equilibrium

We can set up the concentrations at equilibrium in terms of change (x) from initial conditions as follows: \[ H2: [H2]_eq = [H2]_i - x, I2: [I2]_eq = [I2]_i - x, HI: [HI]_eq = [HI]_i + 2x\] According to the balanced chemical equation, 1 mole of H2 or I2 disappearing (left side of equation) will create 2 moles of HI (right side of equation).

Step 4: Set up the Equilibrium Expression and Equation

The equilibrium expression for the reaction based on the Law of Mass Action is: \[ Kc = \frac{[HI]^2}{[H2][I2]} \] From the expression of Step 3, we substitute in the equilibrium values for the concentration variables in the equilibrium expression and get: \[ 54.3 = \frac{([HI_i + 2x])^2}{([H2_i - x]) ([I2_i - x])} \] This is the equation you’ll need to solve for x, which used in the equation from step 3 will give us the equilibrium concentrations.

Key Concepts

Chemical Equilibrium

When a chemical reaction occurs, reactants convert into products, but under certain conditions, the reverse reaction can also proceed where products convert back into reactants. We reach a state called chemical equilibrium when the rate of the forward reaction equals the rate of the reverse reaction, leading to no overall change in concentrations of reactants and products over time. This doesn't mean the reactions have stopped—rather, they continue to occur at an equal rate, creating a dynamic but stable system. Imagine a crowded room where, for every person who leaves, another one enters; even though people are coming and going, the number of people inside the room stays the same. That's how equilibrium works in a chemical reaction.

Law of Mass Action

The Law of Mass Action provides the foundation for understanding how concentrations of reactants and products relate to each other at equilibrium. It states that for a given chemical reaction at equilibrium and constant temperature, the ratio of the product of the concentrations of the products raised to the power of their coefficients in the balanced equation to the product of the concentrations of the reactants raised to the power of their coefficients is constant. This ratio is termed the equilibrium constant, denoted as K_c for concentrations in molarity (mol/L). The law helps us to write equilibrium expressions and thus to calculate concentrations of substances involved in a reaction at equilibrium.

Equilibrium Concentrations

Determining equilibrium concentrations involves mathematically solving for the concentrations of all reactants and products at equilibrium. This process typically entails setting up an equilibrium expression from the balanced chemical equation, determining the initial concentrations, and considering the change in concentration as the system shifts towards equilibrium. This change is often represented by a variable, like x, and it is linked to the stoichiometry of the equation. For the equation H₂ + I₂ ⇌ 2HI, the change in H₂ and I₂ concentrations would decrease by x, while the HI concentration would increase by 2x, indicating how the substances relate to each other when the reaction proceeds.

Molarity

One key metric in chemistry is molarity, denoted as M. It's a way to express the concentration of a solute in a solution. To calculate molarity, you divide the number of moles of the solute by the volume of the solution in liters. Expressed as a formula, it looks like this: Molarity (M) = Number of moles of solute / Volume of solution in liters. In the context of equilibrium, understanding molarity is crucial because it allows us to quantify the concentration of reactants and products, which we need to know in order to use the equilibrium constant expression effectively.