Question

A sample of solid ammonium chloride was placed in an evacuated container and then heated so that it decomposed to ammonia gas and hydrogen chloride gas. After heating, the total pressure in the container was found to be 4.4 atm. Calculate \(K_{\mathrm{p}}\) at this temperature for the decomposition reaction $$\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$$,

Short answer

The short answer based on the step-by-step solution is as follows: For the decomposition reaction \(\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{HCl}(g)\), the partial pressures of ammonia and hydrogen chloride gases are found to be 2.2 atm each. The expression for the equilibrium constant \(K_p\) is given by \(K_p = [\mathrm{NH}_3(g)] \cdot [\mathrm{HCl}(g)]\). Substituting the partial pressures of ammonia and hydrogen chloride, we find \(K_p = (2.2 \: \mathrm{atm}) \cdot (2.2 \: \mathrm{atm}) = 4.84 \: \mathrm{atm^2}\). Therefore, the equilibrium constant \(K_p\) for this decomposition reaction is \(4.84 \: \mathrm{atm^2}\).

Step-by-step solution

Write the balanced equation for the decomposition reaction.

We are given the balanced equation for the decomposition reaction as \[ \mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \]

Calculate the partial pressures of the gases involved.

The pressure in the container after the reaction reached equilibrium is given as 4.4 atm, which is the sum of the partial pressures of ammonia and hydrogen chloride gases. As one mole of ammonia is produced for each mole of hydrogen chloride, their partial pressures must be equal. Let's denote the partial pressure of ammonia and hydrogen chloride gases as \(x\). Then, we have \( 2x = 4.4 \: \mathrm{atm} \) Now solve for \(x\): \( x = \frac{4.4 \: \mathrm{atm}}{2} = 2.2 \: \mathrm{atm} \) Therefore, the partial pressures of ammonia and hydrogen chloride are 2.2 atm each.

Write the expression for the equilibrium constant \(K_p\).

The equilibrium constant \(K_p\) can be expressed as the product of the partial pressures of the product gases divided by the partial pressure of the reactant. However, since the reactant is a solid, it does not have a partial pressure and can be treated as 1. Thus, the expression for \(K_p\) can be written as \[ K_p = [\mathrm{NH}_3(g)] \cdot [\mathrm{HCl}(g)] \]

Use the partial pressures obtained in Step 2 to calculate \(K_p\).

Substitute the partial pressures of ammonia and hydrogen chloride (2.2 atm each) into the expression for \(K_p\), then calculate the value of \(K_p\) as follows: \[ K_p = (2.2 \: \mathrm{atm}) \cdot (2.2 \: \mathrm{atm}) = 4.84 \: \mathrm{atm^2} \] Thus, the equilibrium constant \(K_p\) for the given decomposition reaction is 4.84 atm\(^2\).

Key concepts

Chemical Equilibrium

Imagine a dance where both partners move in harmony with each other, dancing forwards and backwards with no one ever taking the lead. This dance is a metaphor for chemical equilibrium, an important concept in chemistry that occurs when the forward and reverse reactions in a chemical process happen at equal rates. I think of it as a perfectly balanced scale, with the reactants and products reaching a state where their concentrations remain constant over time, even if they are not equal.

Understanding equilibrium involves recognizing that it is a dynamic process—chemical reactions continue to occur, but since the rates are equal, there are no net changes in the concentrations of the reactants and products. This balanced state can be described by an equilibrium constant, symbolized as K for reactions based on concentration, or as Kp for reactions involving gases and their partial pressures. The exact value of the equilibrium constant at a given temperature provides insight into the position of equilibrium and tells us the relative amounts of products and reactants present in the mixture.

Partial Pressures

When gases mix, like in a balloon or our atmosphere, each one exerts a pressure as if it were alone in the container—this is what we refer to as partial pressure. It’s as if each gas carries a part of the total pressure burden. This pressure contributed by a single type of gas in a mixture is proportionate to its mole fraction—in simple terms, its share of the total number of gas particles present.

Diving deeper into this concept, Dalton’s Law of Partial Pressures comes into play, stating that the total pressure of a gas mixture is the sum of the partial pressures of each individual component. In a reaction where gases are produced or consumed, the partial pressures can tell you a lot about the reaction's progression and, as we already know, they're crucial in calculating the equilibrium constant Kp in terms of pressure. Always keep in mind the power of this law when unraveling the mysteries of gas-related equilibrium!

Decomposition Reaction

When you hear the word 'decomposition', think of breaking down or splitting up. In chemistry, a decomposition reaction is a process where a single compound breaks down into two or more simpler substances. This can happen through various means like heating, as seen with our friend ammonium chloride turning into ammonia and hydrogen chloride.

These reactions can be simple, just like our example, or become quite complex, leading to multiple products and intricate equilibrium scenarios. What's captivating about these reactions is how they reflect nature's tendency to break down complex structures into simpler ones. Remember, in decomposition, complex molecules are effectively 'undoing' their formation process, which serves as an essential cycle in chemical reactions and the broader ecosystem.