Problem 101 - General Chemistry: Principles and Modern Applications

Question

In the reversible reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons\) \(2 \mathrm{HI}(\mathrm{g}),\) an initial mixture contains \(2 \mathrm{mol} \mathrm{H}_{2}\) and 1 mol I \(_{2} .\) The amount of HI expected at equilibrium is (a) \(1 \mathrm{mol} ;\) (b) \(2 \mathrm{mol} ;\) (c) less than \(2 \mathrm{mol}\); (d) more than 2 mol but less than 4 mol.

Step-by-Step Solution

Answer

The correct option explaining the amount of HI expected at equilibrium is option (b), which states it would be 2 mol.

Step 1: Understand the Reversible Reaction

First, comprehend that the reaction is reversible indicating that both the forward (reactants forming products) and backward (products forming reactants) reactions are occurring concurrently when the system is at equilibrium. The equation for this reaction is: \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\). The reactants are hydrogen and iodine, which form hydrogen iodide in the product.

Step 2: Determine the Initial Amount of Each Species

As per the exercise, the initial amounts are 2 mol of hydrogen (\(H_{2}\)) and 1 mol of iodine (\(I_{2}\)). At the start, no hydrogen iodide is present, hence its initial amount is 0 mol.

Step 3: Evaluate the Stoichiometry and Equilibrium

From the balanced equation, for every one mol of \(H_2\) and \(I_2\) that react, they form 2 mol of \(HI\). However, initially there is an excess of hydrogen which means that all 1 mol of \(I_2\) will be consumed, producing 2 mol of \(HI\) in the process. After this, no more \(HI\) can be produced as \(I_2\) is now the limiting reactant.

Step 4: Select the right answer

From the previous steps, it is established that only 2 mol of \(HI\) can be produced. This aligns with option (b).

Key Concepts

Reversible Reaction

In chemistry, a reversible reaction is one where the reactants form products, and simultaneously, the products can revert to the reactants. Such reactions are represented with a double-headed arrow in their equation:
\[\text{Reactants} \rightleftharpoons \text{Products}\]
For the reaction \( \mathrm{H}_{2} + \mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI} \), both the formation of \( \mathrm{HI} \) from \( \mathrm{H}_{2} \) and \( \mathrm{I}_{2} \), and the breakdown of \( \mathrm{HI} \) back into \( \mathrm{H}_{2} \) and \( \mathrm{I}_{2} \), occur at the same time.

• At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.
• The concentrations of reactants and products remain constant, though they may not necessarily be equal.

Understanding reversible reactions is crucial because many chemical reactions in biological, chemical, and industrial processes are reversible, meaning they reach a state of dynamic balance.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in chemical reactions. It allows us to predict how much product we can expect from a given amount of reactants, assuming complete reaction according to the balanced chemical equation.
In the reaction \( \mathrm{H}_{2} + \mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI} \), the stoichiometric coefficients can be interpreted as ratios:

• 1 mole of \( \mathrm{H}_{2} \) reacts with 1 mole of \( \mathrm{I}_{2} \).


• This produces 2 moles of \( \mathrm{HI} \).


• If you start with 2 moles of \( \mathrm{H}_{2} \) and 1 mole of \( \mathrm{I}_{2} \), theoretically you should end with 2 moles of \( \mathrm{HI} \), because \( \mathrm{I}_{2} \) is the limiting factor.

Stoichiometry is essential for calculating the quantities of substances needed or produced in a given chemical reaction. It ensures you use resources efficiently, reducing waste and cost.

Limiting Reactant

The concept of a limiting reactant is pivotal when conducting reactions with specific initial quantities of each reactant. The limiting reactant is the substance that is completely used up first halting the reaction until more of that reactant is added. This reactant limits the amount of product that can be formed in the reaction.
For the given reaction \( \mathrm{H}_{2} + \mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI} \), if we start with 2 moles of \( \mathrm{H}_{2} \) and 1 mole of \( \mathrm{I}_{2} \), \( \mathrm{I}_{2} \) becomes the limiting reactant because:

• Once the 1 mole of \( \mathrm{I}_{2} \) is consumed, no additional \( \mathrm{HI} \) can be produced due to the lack of \( \mathrm{I}_{2} \).


• This results in a maximum yield of 2 moles of \( \mathrm{HI} \).

Understanding which reactant is the limiting reactant is important because it helps in predicting the maximum possible amount of product and understanding the constraints posed by the available reactants.