Problem 25 - Chemistry The Molecular Science

Question

The reaction of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) with chlorine trifluoride, \(\mathrm{ClF}_{3}\), has been used in experimental rocket motors. \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+\frac{4}{3} \mathrm{ClF}_{3}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{HF}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{~g})+\frac{2}{3} \mathrm{Cl}_{2}(\mathrm{~g})\) How is the equilibrium constant, \(K_{\mathrm{p}}\), for this reaction related to \(K_{\mathrm{p}}^{\prime}\) for the reaction written this way? $$ \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+4 \mathrm{ClF}_{3}(\mathrm{~g}) \rightleftharpoons 12 \mathrm{HF}(\mathrm{g})+3 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ (a) \(K_{\mathrm{P}}=K_{\mathrm{P}}^{\prime}\) (b) \(K_{\mathrm{P}}=1 / K_{\mathrm{P}}^{\prime}\) (c) \(K_{\mathrm{p}}^{3}=K_{\mathrm{P}}^{\prime}\) (d) \(K_{\mathrm{P}}=\left(K_{\mathrm{P}}^{\prime}\right)^{3}\) (e) \(3 K_{\mathrm{p}}=K_{\mathrm{P}}^{\prime}\)

Step-by-Step Solution

Answer

Option (c): \(K_{p}^{3} = K_{p}^{\prime}\).

Step 1: Analyze Reaction Stoichiometry

Start by analyzing the stoichiometric coefficients of the gases involved in the reaction. The first equation has: 1 mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\), \(\frac{4}{3}\) moles of \(\mathrm{ClF}_{3}\), producing 4 moles of \(\mathrm{HF}\), 1 mole of \(\mathrm{N}_{2}\), and \(\frac{2}{3}\) moles of \(\mathrm{Cl}_{2}\). The second equation has 1 mole of \(\mathrm{N}_{2} \mathrm{H}_{4}\), 4 moles of \(\mathrm{ClF}_{3}\), producing 12 moles of \(\mathrm{HF}\), 3 moles of \(\mathrm{N}_{2}\), and 1 mole of \(\mathrm{Cl}_{2}\).

Step 2: Compare Reaction Ratios

Observe that the second reaction is a factor of 3 times the first reaction. Each product in the second reaction has coefficients exactly 3 times larger than in the first reaction.

Step 3: Determine Relation between Equilibrium Constants

The relationship between the equilibrium constants when reactions are scaled by a factor is \(K^n\) where \(n\) is the scaling factor. In this case, each term is scaled by 3, so \(K_{p}^{\prime} = (K_{p})^{3}\).

Step 4: Select Correct Answer

Based on the determined relation, the correct answer is the one that fits \(K_{p}^{3} = K_{p}^{\prime}\). Thus, option (c) \(K_{p}^{3}=K_{p}^{\prime}\) is correct.

Key Concepts

Chemical Equilibrium

In chemical reactions, equilibrium is the state where the reactants and products have reached a balance. When this state is achieved, the concentrations of all species remain constant over time. This doesn't mean the reaction has stopped but rather that the forward and backward reactions occur at the same rate.
The equilibrium constant, often denoted as \(K\), provides a numerical value reflecting this balance. For reactions involving gases, the equilibrium constant is expressed as \(K_p\), using partial pressures.

• A large \(K_p\) indicates a reaction favoring products at equilibrium.
• A small \(K_p\) suggests a reaction favoring reactants.

Understanding equilibrium helps in predicting the direction of a reaction under certain conditions, which is crucial in fields like chemistry and biochemistry.
In the given examples, the relation between two equilibrium constants was derived using the stoichiometric adjustments of the reactions, demonstrating how changes in equation coefficients affect \(K_p\).

Stoichiometry

Stoichiometry involves quantitative relationships between reactants and products in a chemical reaction. It ensures that matter is conserved according to the balanced equation of the reaction. In the given problem, two different representations of the reaction were provided.
Comparing the stoichiometric coefficients from both equations, it's clear that the second reaction's coefficients are three times those of the first.

• This implies a simple scaling relationship between the two.
• Each molecule in the second equation exists in a quantity three times that in the first.

When stoichiometry changes in this manner, the equilibrium constant changes accordingly. The relationship is dictated by the scaling factor, exemplifying a core principle in stoichiometry: coefficients impact the calculation of equilibrium constants and the proportions of reactants and products.

Chemical Reactions

Chemical reactions describe the process where reactants transform into products, altering bonds and creating new substances. These reactions can be influenced by a variety of factors, including temperature, pressure, and concentration of reactants.
In the context of the exercise, two chemical reactions involving hydrazine and chlorine trifluoride were analyzed.
The reactions generate products like hydrogen fluoride, nitrogen gas, and chlorine gas.

• Each variant of the reaction has different stoichiometric coefficients, impacting the equilibrium expression.
• The importance of substitutes also affects reaction equilibria values like \(K_p\).

By altering the coefficients in the chemical equation, the exponents in the equilibrium expression are similarly affected, illustrating how sensitivity analysis is important in chemistry for scaling reactions appropriately. This shift underlines the dynamism intrinsic to chemical reactions.