A homogeneous ideal gaseous reaction, AB2(g) A(g) + 2B(g) is carried out in a 25 L flask at 27°C

Q: Why do we use $PV=nRT$ at equilibrium?

Because the mixture of gases at equilibrium still behaves as an ideal gas. The total pressure is exerted by the total moles of all gases present in the flask at that specific temperature and volume.

Q: Why is the change for B $+2x$ and not just $+x$?

Stoichiometry! For every 1 mole of $AB_2$ that decomposes, 2 moles of $B$ are produced according to the balanced chemical equation.

Q: Can we calculate $K_c$ first and then convert to $K_p$?

Yes, you could find the molar concentrations, calculate $K_c$, and then use $K_p = K_c(RT)^{\Delta n_g}$. However, since total pressure is given, working directly with partial pressures is often more straightforward.

Q: Why do we square the partial pressure of B in the $K_p$ expression?

The Law of Mass Action states that the partial pressure of each species must be raised to the power of its stoichiometric coefficient from the balanced equation. Since B has a coefficient of 2, its partial pressure is squared.

Q: What if the volume of the flask was changed?

According to Le Chatelier's Principle, decreasing the volume increases the pressure, shifting the equilibrium towards the side with fewer moles of gas (in this case, backwards). The value of $K_p$ itself, however, remains constant as long as the temperature doesn't change.

Question

A homogeneous ideal gaseous reaction, $AB_2(g) ightleftharpoons A(g) + 2B(g)$ is carried out in a 25 L flask at 27°C. The initial amount of $AB_2$ was 1 mole and the equilibrium pressure was 1.9 atm. The value of $K_p$ is $x 	imes 10^{-2}$. The value of $x$ is ………… .

Sigma Percentile · Accepted Answer

### The Setup: A Flask Full of Potential

Imagine a $25\,	ext{L}$ flask, sealed and sitting at a comfortable $27^\circ	ext{C}$ (which is $300\,	ext{K}$). Inside, we initially place exactly $1\,	ext{mole}$ of a gaseous compound, $AB_2$. As time passes, this gas begins to decompose, establishing a dynamic equilibrium:

$$AB_2(g) ightleftharpoons A(g) + 2B(g)$$

We are given a crucial piece of information: once the system settles into equilibrium, the total pressure inside the flask is $1.9\,	ext{atm}$. Our mission is to find the equilibrium constant, $K_p$, and express it in the form $x 	imes 10^{-2}$.

### The ICE Table: Tracking the Change

To understand what's happening inside the flask, we must track the moles of each gas. We do this using an **ICE (Initial, Change, Equilibrium)** table. Let's assume that $x$ moles of $AB_2$ dissociate to reach equilibrium.

Initially, we have $1\,	ext{mole}$ of $AB_2$ and $0\,	ext{moles}$ of $A$ and $B$.

As the reaction proceeds, $AB_2$ loses $x$ moles. According to the stoichiometry of the balanced equation, for every $1\,	ext{mole}$ of $AB_2$ that decomposes, $1\,	ext{mole}$ of $A$ and $2\,	ext{moles}$ of $B$ are formed. Therefore, the change is $+x$ for $A$ and $+2x$ for $B$.

At equilibrium, the moles are:
- $n_{AB_2} = 1 - x$
- $n_A = x$
- $n_B = 2x$

The total number of moles at equilibrium, $n_{	ext{total}}$, is simply the sum of these:

$$n_{	ext{total}} = (1 - x) + x + 2x = 1 + 2x$$

### The Ideal Gas Bridge: Finding the Missing Link

We have an expression for the total moles, but we need a numerical value for $x$. This is where the **Ideal Gas Law** comes to our rescue. Since we know the total equilibrium pressure ($P = 1.9\,	ext{atm}$), the volume ($V = 25\,	ext{L}$), and the temperature ($T = 300\,	ext{K}$), we can find the total moles.

$$PV = n_{	ext{total}}RT$$

Substituting our known values:

$$1.9 	imes 25 = (1 + 2x) 	imes 0.082 	imes 300$$

Solving for the term $(1 + 2x)$:

$$1 + 2x = \frac{1.9 	imes 25}{0.082 	imes 300} = \frac{47.5}{24.6} \approx 1.93$$

Now, we can easily isolate $x$:

$$2x = 0.93 \implies x = 0.465$$

### Dalton's Law and the Equilibrium Constant

To calculate $K_p$, we need the partial pressure of each gas. According to **Dalton's Law of Partial Pressures**, the partial pressure of a gas is its mole fraction multiplied by the total pressure ($p_i = \chi_i P$).

- $p_{AB_2} = \left(\frac{1-x}{1+2x}ight)P$
- $p_A = \left(\frac{x}{1+2x}ight)P$
- $p_B = \left(\frac{2x}{1+2x}ight)P$

The expression for the equilibrium constant $K_p$ is the product of the partial pressures of the products divided by the partial pressure of the reactants, each raised to the power of their stoichiometric coefficients:

$$K_p = \frac{p_A 	imes p_B^2}{p_{AB_2}}$$

Substituting our partial pressure expressions into the $K_p$ equation:

$$K_p = \frac{\left(\frac{x}{1+2x}Pight) \left(\frac{2x}{1+2x}Pight)^2}{\left(\frac{1-x}{1+2x}Pight)}$$

Simplifying this algebraically gives us a much cleaner formula to work with:

$$K_p = \frac{4x^3 P^2}{(1+2x)^2 (1-x)}$$

### The Final Calculation: Patience is Key

Now, we carefully substitute our numerical values: $x = 0.465$, $P = 1.9$, and $(1+2x) = 1.93$.

$$K_p = \frac{4(0.465)^3 (1.9)^2}{(1.93)^2 (1-0.465)}$$

$$K_p = \frac{4 	imes 0.1005 	imes 3.61}{3.7249 	imes 0.535}$$

$$K_p = \frac{1.451}{1.9928} \approx 0.7285\,	ext{atm}$$

The question asks for the value in the format $x 	imes 10^{-2}$.

$$0.7285 = 72.85 	imes 10^{-2} \approx 73 	imes 10^{-2}$$

Therefore, our final integer answer is **73**.

Enter Numerical Value:

Visualized Solution

$A B_{2} (g) ⇌ A (g) + 2 B (g)$

$I C E$

$n_{t o t a l}$

$P V = n R T$

$x$

$p_{i}$

$K_{p}$

$K_{p} (x, P)$

$K_{p}$

$x$

$Δ V$

The Sigma Insight: Law of Mass Action

The Setup

The ICE Table

The Ideal Gas Bridge

Dalton's Law and the Equilibrium Constant

The Final Calculation

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Consider the reaction, $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ The equilibrium constant of the above reaction is $K_{p}$ . If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that $p_{N H_{3}} ≪ p_{total}$ at equilibrium)

Comprehension Passage

The equilibrium constant $K_{P}$ for this reaction at 298 K, in terms of $β_{equilibrium}$ , is

The INCORRECT statement among the following , for this reaction, is

For the reaction, $A + B ⇌ 2 C$ The value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is $x \times 1 0^{- 1}$ M. The value of $x$ is ...... . (Nearest integer)

For the reaction, $2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g)$ In an equilibrium mixture, the partial pressures are $p_{SO_{3}} = 43 kPa$ ; $p_{O_{2}} = 530 Pa$ and $p_{SO_{2}} = 45 kPa$ . The equilibrium constant $K_{p} = \dots \times 1 0^{- 2}$ . (Nearest integer)

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB2​(g)⇌A(g)+2B(g)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ICE

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ntotal​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PV=nRT

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pi​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kp​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kp​(x,P)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kp​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔV

The Sigma Insight: Law of Mass Action

The Setup

The ICE Table

The Ideal Gas Bridge

Dalton's Law and the Equilibrium Constant

The Final Calculation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equilibrium constant KP​ for this reaction at 298 K, in terms of βequilibrium​ , is

The INCORRECT statement among the following , for this reaction, is

$A B_{2} (g) ⇌ A (g) + 2 B (g)$

$I C E$

$n_{t o t a l}$

$P V = n R T$

$x$

$p_{i}$

$K_{p}$

$K_{p} (x, P)$

$K_{p}$

$x$

$Δ V$

The equilibrium constant $K_{P}$ for this reaction at 298 K, in terms of $β_{equilibrium}$ , is