A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on

Q: Why don't we include the pressure of solid graphite in our total pressure calculation?

Solid graphite does not exert any vapor pressure under these conditions. Total pressure in the vessel is purely due to the collisions of gas molecules ($\text{CO}_2$ and $\text{CO}$) with the walls.

Q: Why is the change for $\text{CO}$ written as $+2p$ instead of just $+p$?

Look at the balanced equation: $\text{CO}_2(g) + \text{C}(s) \rightleftharpoons 2\text{CO}(g)$. For every 1 mole of $\text{CO}_2$ that reacts, 2 moles of $\text{CO}$ are produced. Hence, if $\text{CO}_2$ pressure drops by $p$, $\text{CO}$ pressure must increase by $2p$.

Q: Why is the concentration or pressure of solid carbon ignored in the $K_p$ expression?

In equilibrium expressions, the activity of pure solids and pure liquids is taken as unity (1). Their concentration remains constant throughout the reaction, so they are incorporated into the equilibrium constant itself.

Q: What does the value of $p$ physically represent?

The value $p$ represents the drop in the partial pressure of $\text{CO}_2$ as it gets consumed to reach equilibrium. It's a direct measure of how far the reaction has proceeded in the forward direction.

Q: Can we use concentrations instead of partial pressures to solve this?

Since the volume and temperature are constant, partial pressure is directly proportional to the number of moles (and thus concentration). However, since the problem gives data in atm and asks for $K_p$, it is much more direct and less error-prone to work entirely in partial pressures.

Question

A vessel at $1000	ext{ K}$ contains $	ext{CO}_2$ with a pressure of $0.5	ext{ atm}$. Some of the $	ext{CO}_2$ is converted into $	ext{CO}$ on the addition of graphite. If the total pressure at equilibrium is $0.8	ext{ atm}$, the value of $K_p$ is

Options:

(A) $1.8	ext{ atm}$

(B) $3	ext{ atm}$

(C) $0.3	ext{ atm}$

(D) $0.18	ext{ atm}$

Sigma Percentile · Accepted Answer

### Setting the Stage: The Closed Vessel

Imagine you are observing a closed reaction vessel at a scorching $1000	ext{ K}$. Inside, we have a block of solid graphite sitting quietly at the bottom, surrounded by a swirling cloud of carbon dioxide ($	ext{CO}_2$) gas.

The initial pressure of this $	ext{CO}_2$ gas is given as $0.5	ext{ atm}$.

This is our starting point. The system is not yet at equilibrium, and a chemical transformation is about to begin.

### The Stoichiometric Dance: Setting up the ICE Table

As the reaction proceeds, the $	ext{CO}_2$ gas reacts with the solid carbon to produce carbon monoxide ($	ext{CO}$) gas.

The balanced chemical equation is:
$$ 	ext{CO}_2(g) + 	ext{C}(s) ightleftharpoons 2	ext{CO}(g) $$

Notice the stoichiometry here. For every one mole of $	ext{CO}_2$ that reacts, two moles of $	ext{CO}$ are produced. This is a crucial detail!

To track the changes, we set up an **ICE (Initial, Change, Equilibrium)** table. Let's assume the partial pressure of $	ext{CO}_2$ decreases by an amount $p$ to reach equilibrium.

Therefore, the equilibrium pressure of $	ext{CO}_2$ becomes $(0.5 - p)	ext{ atm}$.

Consequently, the pressure of the newly formed $	ext{CO}$ gas will be $+2p	ext{ atm}$.

### The Pressure Puzzle: Finding the Unknown

We are given a vital piece of information: the total pressure at equilibrium is $0.8	ext{ atm}$.

But what contributes to this total pressure? **Only the gases exert pressure.** The solid graphite does not contribute to the gas pressure.

So, the total pressure is simply the sum of the partial pressures of $	ext{CO}_2$ and $	ext{CO}$:
$$ P_{	ext{total}} = P_{	ext{CO}_2} + P_{	ext{CO}} $$

Substituting our equilibrium expressions:
$$ 0.8 = (0.5 - p) + 2p $$

Solving this simple linear equation, we find:
$$ 0.8 = 0.5 + p $$
$$ p = 0.3	ext{ atm} $$

### The Final Act: Calculating the Equilibrium Constant

Now that we have the value of $p$, we can find the exact equilibrium partial pressures of our gases.

For $	ext{CO}_2$:
$$ P_{	ext{CO}_2} = 0.5 - 0.3 = 0.2	ext{ atm} $$

For $	ext{CO}$:
$$ P_{	ext{CO}} = 2(0.3) = 0.6	ext{ atm} $$

Finally, we write the expression for the equilibrium constant, $K_p$. Remember, **the activity of pure solids is taken as unity (1)**, so solid carbon does not appear in the expression.

$$ K_p = \frac{(P_{	ext{CO}})^2}{P_{	ext{CO}_2}} $$

Plugging in our calculated values:
$$ K_p = \frac{(0.6)^2}{0.2} $$
$$ K_p = \frac{0.36}{0.2} = 1.8	ext{ atm} $$

And there we have it! The equilibrium constant $K_p$ for this reaction is $1.8	ext{ atm}$.

Animated Solution for Chemistry - Chemical Equilibrium: 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

Select Answer:

Visualized Solution

\text{Initial Setup}

\text{Chemical Equation}

\text{ICE Table}

\text{Total Pressure}

\text{Solving for } p

\text{Equilibrium Partial Pressures}

\text{Equilibrium Constant } K_p

\text{Final Calculation}

The Sigma Insight: Law of Mass Action

Setting the Stage

The Stoichiometric Dance

The Pressure Puzzle

The Final Act

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