Consider the reaction, N2(g) + 3H2(g) 2NH3(g) The equilibrium constant of the above reaction is

Q: Why did we use the formation reaction for $K_p$ instead of the dissociation reaction?

The problem explicitly states that $K_p$ is the equilibrium constant for the 'above reaction', which is written as $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$. We must always follow the exact chemical equation provided when writing the equilibrium expression.

Q: Why is $p_{N_2} = P$ and $p_{H_2} = 3P$?

When pure ammonia dissociates, it follows the stoichiometry of the reaction: $2NH_3 \rightarrow N_2 + 3H_2$. For every 1 mole of $N_2$ produced, 3 moles of $H_2$ are produced. Since partial pressure is directly proportional to the number of moles (at constant volume and temperature), their pressures will be in a $1:3$ ratio.

Q: How does the approximation $p_{NH_3} \ll p_{\text{total}}$ actually help us?

The total pressure is $p = p_{N_2} + p_{H_2} + p_{NH_3}$. Without the approximation, we would have $p = 4P + p_{NH_3}$, which leaves us with two unknowns ($P$ and $p_{NH_3}$) in our total pressure equation. By assuming $p_{NH_3}$ is negligible, we simplify it to $p \approx 4P$, allowing us to easily express $P$ in terms of $p$.

Q: What if the approximation wasn't given? Could we still solve it?

Yes, but it would be much more mathematically rigorous. You would have to substitute $P = \frac{p - p_{NH_3}}{4}$ into the $K_p$ expression, which would lead to a high-order polynomial equation for $p_{NH_3}$ that is very difficult to solve analytically.

Q: Why did $27$ become $3^{3/2}$ in the final answer?

In the equation, we had $27$ in the numerator for $p_{NH_3}^2$. We can write $27$ as $3^3$. When we take the square root of both sides to find $p_{NH_3}$, the square root of $3^3$ is $(3^3)^{1/2}$, which simplifies to $3^{3/2}$.

Question

Consider the reaction,
$N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$
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 $p_{NH_3} \ll p_{	ext{total}}$ at equilibrium)

Options:

(A) $\frac{3^{3/2} K_p^{1/2} p^2}{4}$

(B) $\frac{3^{3/2} K_p^{1/2} p^2}{16}$

(C) $\frac{K_p^{1/2} p^2}{16}$

(D) $\frac{K_p^{1/2} p^2}{4}$

Sigma Percentile · Accepted Answer

### The Setup: A Dance of Molecules

Imagine a closed vessel where pure ammonia gas ($NH_3$) is introduced and allowed to reach equilibrium. The ammonia molecules begin to dissociate, breaking apart to form nitrogen ($N_2$) and hydrogen ($H_2$) gases. The chemical equation for this dissociation is:

$$2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$$

However, there is a slight twist in the problem statement. The equilibrium constant $K_p$ provided is for the *formation* of ammonia, not its dissociation. The formation reaction is written as:

$$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$

This means when we write our expression for $K_p$, we must strictly follow the stoichiometry of the formation reaction.

### The Power of Approximation

Let's analyze the partial pressures at equilibrium. Since the dissociation of ammonia produces 1 mole of $N_2$ for every 3 moles of $H_2$, their partial pressures will always be in a $1:3$ ratio. Let's define the partial pressure of nitrogen as $P$. Consequently, the partial pressure of hydrogen will be $3P$.

$$p_{N_2} = P$$
$$p_{H_2} = 3P$$

The total pressure in the vessel, let's call it $p$, is the sum of the partial pressures of all the gases present:

$$p = p_{N_2} + p_{H_2} + p_{NH_3}$$

Here comes the crucial, time-saving approximation given in the problem: the partial pressure of ammonia is extremely small compared to the total pressure ($p_{NH_3} \ll p$). This allows us to safely ignore $p_{NH_3}$ when calculating the total pressure.

$$p \approx p_{N_2} + p_{H_2}$$
$$p \approx P + 3P = 4P$$

From this, we can easily express our assumed variable $P$ in terms of the total pressure $p$:

$$P = \frac{p}{4}$$

### The Master Equation

Now, let's construct the equilibrium constant expression. Remember, we must use the formation reaction for which $K_p$ is defined:

$$K_p = \frac{p_{NH_3}^2}{p_{N_2} \cdot p_{H_2}^3}$$

We substitute our expressions for $p_{N_2}$ and $p_{H_2}$ into this equation:

$$K_p = \frac{p_{NH_3}^2}{P \cdot (3P)^3}$$
$$K_p = \frac{p_{NH_3}^2}{27P^4}$$

### The Final Algebraic Sprint

We are almost there! We need to find the partial pressure of ammonia, $p_{NH_3}$. Let's substitute $P = \frac{p}{4}$ into our $K_p$ equation:

$$K_p = \frac{p_{NH_3}^2}{27 \left(\frac{p}{4}\right)^4}$$
$$K_p = \frac{p_{NH_3}^2 \cdot 4^4}{27 p^4}$$

Now, we rearrange the equation to isolate $p_{NH_3}^2$:

$$p_{NH_3}^2 = \frac{27 p^4 K_p}{4^4}$$

Recognizing that $27$ is $3^3$, we can write:

$$p_{NH_3}^2 = \frac{3^3 p^4 K_p}{4^4}$$

Finally, we take the square root of both sides to find $p_{NH_3}$:

$$p_{NH_3} = \frac{3^{3/2} K_p^{1/2} p^2}{16}$$

And there we have it! By carefully applying stoichiometry and leveraging a powerful approximation, we've arrived at the correct expression.

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Visualized Solution

$2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$

$Partial Pressures$

$Total Pressure$

$Evaluating P$

$Equilibrium Constant K_{p}$

$Substituting Pressures$

$Substituting P = \frac{p}{4}$

$Solving for p_{N H_{3}}^{2}$

$Final Expression$

$Without Approximation$

The Sigma Insight: Law of Mass Action

The Setup

The Power of Approximation

The Master Equation

The Final Algebraic Sprint

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2NH3​(g)⇌N2​(g)+3H2​(g)

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting P=4p​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for pNH3​2​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Without Approximation