Two solids dissociate as follows: A(s) B(g) + C(g); K_p1 = x atm^2 D(s) C(g) + E(g); K_p2 =

Q: Why don't we include the solids A and D in the $K_p$ expression?

The active mass (or concentration/partial pressure) of a pure solid is taken as unity (1) because its density remains constant during the reaction. It does not exert a partial pressure.

Q: Why is the partial pressure of C taken as $(p_1 + p_2)$ in both equations?

Because both reactions occur in the same vessel, the gas C molecules from both sources mix together. The equilibrium for each reaction depends on the *total* partial pressure of C present in the system, not just the C it produced itself.

Q: Can we solve for $p_1$ and $p_2$ individually?

Yes, you can divide the two $K_p$ equations to get $p_1/p_2 = x/y$, which allows you to find individual pressures, but it's not necessary for finding the total pressure. Adding them is a much faster shortcut.

Q: What is Dalton's Law of Partial Pressures?

It states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. Here, $P_{total} = p_B + p_C + p_E$.

Q: What if the volume of the vessel was given?

The equilibrium constants $K_p$ are independent of volume. The partial pressures would adjust to maintain the $K_p$ values, so the total pressure would remain the same as long as some solid is still present to maintain equilibrium.

Question

Two solids dissociate as follows:
$A(s) ightleftharpoons B(g) + C(g); K_{p_1} = x 	ext{ atm}^2$
$D(s) ightleftharpoons C(g) + E(g); K_{p_2} = y 	ext{ atm}^2$
The total pressure when both the solids dissociate simultaneously is

Options:

(A) $\sqrt{x+y} 	ext{ atm}$

(B) $x^2 + y^2 	ext{ atm}$

(C) $(x+y) 	ext{ atm}$

(D) $2(\sqrt{x+y}) 	ext{ atm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a closed reaction vessel. Inside, there are two distinct solid substances, $A$ and $D$. When heated, they both begin to decompose into gaseous products simultaneously. This is a classic scenario known as **simultaneous equilibria**.

The reactions are given as:
$$A(s) \rightleftharpoons B(g) + C(g)$$
$$D(s) \rightleftharpoons C(g) + E(g)$$

The most crucial observation here is the presence of **Gas C** in both reactions. This is an application of the common ion effect (or common gas effect). Because both reactions are happening in the same container, the molecules of gas $C$ produced by solid $A$ mix perfectly with the molecules of gas $C$ produced by solid $D$.

### Setting Up the Partial Pressures

Let's assign some variables to make the math manageable. Suppose the dissociation of solid $A$ produces gas $B$ with a partial pressure of $p_1$. Because the stoichiometry between $B$ and $C$ is $1:1$, solid $A$ also produces gas $C$ with a partial pressure of $p_1$.

Similarly, let the dissociation of solid $D$ produce gas $E$ with a partial pressure of $p_2$. It will also produce gas $C$ with a partial pressure of $p_2$.

Now, what is the **total partial pressure** of gas $C$ in the vessel? It's simply the sum of the contributions from both reactions:
$$p_C = p_1 + p_2$$

### The Master Equations

We are given the equilibrium constants for both reactions, $K_{p_1} = x$ and $K_{p_2} = y$. Remember, the active mass of a pure solid is taken as $1$, so we only include the gaseous products in our $K_p$ expressions.

For the first reaction:
$$K_{p_1} = p_B \cdot p_C$$
$$x = p_1(p_1 + p_2)$$

For the second reaction:
$$K_{p_2} = p_E \cdot p_C$$
$$y = p_2(p_1 + p_2)$$

We have a system of two equations. Our goal is to find the total pressure, which will require us to know the sum of the partial pressures. Let's see what happens if we add these two equations together:

$$x + y = p_1(p_1 + p_2) + p_2(p_1 + p_2)$$

Notice that $(p_1 + p_2)$ is a common factor on the right side. Let's factor it out:

$$x + y = (p_1 + p_2)(p_1 + p_2)$$
$$x + y = (p_1 + p_2)^2$$

Taking the square root of both sides gives us a beautiful intermediate result:
$$p_1 + p_2 = \sqrt{x + y}$$

### Final Calculation

The question asks for the **total pressure** in the vessel when both solids dissociate simultaneously. According to Dalton's Law of Partial Pressures, the total pressure is the sum of the partial pressures of all the individual gases present.

$$P_{total} = p_B + p_C + p_E$$

Substitute the variables we defined earlier:
$$P_{total} = p_1 + (p_1 + p_2) + p_2$$

Combine the like terms:
$$P_{total} = 2p_1 + 2p_2 = 2(p_1 + p_2)$$

Now, we just need to substitute our intermediate result into this equation:
**$$P_{total} = 2\sqrt{x + y}$$**

This elegant result shows how interconnected simultaneous equilibria can be. The total pressure isn't just a simple sum of $x$ and $y$; it's fundamentally tied to the square root of their sum due to the shared gas $C$.

Animated Solution for Chemistry - Chemical Equilibrium: Two solids dissociate as follows: $A (s) ⇌ B (g) + C (g); K_{p_{1}} = x atm^{2}$ $D (s) ⇌ C (g) + E (g); K_{p_{2}} = y atm^{2}$ The total pressure when both the solids dissociate simultaneously is

Select Answer:

Visualized Solution

\text{Simultaneous Equilibria}

\text{Partial Pressures of } B \text{ and } E

\text{Common Gas } C

\text{Equilibrium Constants}

\text{Adding the Equations}

\text{Solving for } p_1 + p_2

\text{Total Pressure}

\text{Final Substitution}

\text{The Way Forward}

The Sigma Insight: Law of Mass Action

Analyzing the Setup

Setting Up the Partial Pressures

The Master Equations

Final Calculation

Similar Questions

The equilibrium constants $K_{p_{1}}$ and $K_{p_{2}}$ for the reactions $X ⇌ 2 Y$ and $Z ⇌ P + Q$ , respectively are in the ratio of $1 : 9$ . If the degree of dissociation of $X$ and $Z$ be equal, then the ratio of total pressure at these equilibria is

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel, $PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g)$ If total pressure, at equilibrium, of the reaction mixture is $p$ and degree of dissociation of $PCl_{5}$ is $x$ , the partial pressure of $PCl_{3}$ will be

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)

If the equilibrium constant for $A ⇌ B + C$ is $K_{e q}^{(1)}$ and that of $B + C ⇌ P$ is $K_{e q}^{(2)}$ , the equilibrium constant for $A ⇌ P$ is

Consider the following reversible chemical reactions, $A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$ ...(i) with equilibrium constant $K_{1}$ $6 A B (g) ⇌ 3 A_{2} (g) + 3 B_{2} (g)$ ...(ii) with equilibrium constant $K_{2}$ The relation between $K_{1}$ and $K_{2}$ is

A homogeneous ideal gaseous reaction, $A B_{2} (g) ⇌ A (g) + 2 B (g)$ is carried out in a 25 L flask at 27°C. The initial amount of $A B_{2}$ was 1 mole and the equilibrium pressure was 1.9 atm. The value of $K_{p}$ is $x \times 1 0^{- 2}$ . The value of $x$ is ………… .

A vessel at $1000 K$ contains $CO_{2}$ with a pressure of $0.5 atm$ . Some of the $CO_{2}$ is converted into $CO$ on the addition of graphite. If the total pressure at equilibrium is $0.8 atm$ , the value of $K_{p}$ is