Consider the following equilibrium:2

NH

3

g

  +

N

2

g

3

H

2

g

=

 

 

 

 

 

 

 

Δ

G

0

34.

kJ

Now suppose a reaction vessel is filled with 4.48

atm

 of ammonia NH

3

 and 0.628

atm

 of hydrogen H

2

 at 333.

°C

. Answer the following questions about this system:

Under these conditions, will the pressure of H

2

 tend to rise or fall?

rise



fall

Is it possible to reverse this tendency by adding N

2

?

In other words, if you said the pressure of H

2

 will tend to rise, can that be changed to a tendency to fall by adding N

2

? Similarly, if you said the pressure of H

2

 will tend to fall, can that be changed to a tendency to rise by adding N

2

?

yes



no

If you said the tendency can be reversed in the second question, calculate the minimum pressure of N

2

 needed to reverse it.

Round your answer to 2

 significant digits.atm

Let's go step-by-step to answer your questions based on the provided chemical equilibrium system.

### Given equilibrium:

\[
\text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\text{H}_2(g) \quad \Delta G^0 = 34 \, \text{kJ/mol}
\]

### Step 1: Reaction quotient (Q) and equilibrium constant (K)

1. **ΔG and K:**
   The relationship between the standard Gibbs free energy change (\( \Delta G^0 \)) and the equilibrium constant (\( K \)) is given by the formula:

   \[
   \Delta G^0 = -RT \ln K
   \]

   Where:
   - \( R \) = 8.314 J/mol·K (gas constant),
   - \( T \) is the temperature in Kelvin,
   - \( K \) is the equilibrium constant.

   First, convert the temperature to Kelvin:
   \[
   T = 333 + 273.15 = 606.15 \, K
   \]

   Now solve for \( K \):

   \[
   34,000 = - (8.314 \times 606.15) \ln K
   \]

   \[
   \ln K = \frac{-34,000}{(8.314 \times 606.15)} = -6.81
   \]

   \[
   K = e^{-6.81} \approx 0.0011
   \]

### Step 2: Reaction Quotient (Q)

Now let's calculate the reaction quotient \( Q \), which will help us determine the direction in which the reaction will shift.

\[
Q = \frac{P_{\text{N}_2} P_{\text{H}_2}^3}{P_{\text{NH}_3}}
\]

Given:
- \( P_{\text{NH}_3} = 4.48 \, \text{atm} \),
- \( P_{\text{H}_2} = 0.628 \, \text{atm} \).

We don't know the partial pressure of nitrogen, but for now, let’s assume the pressure of nitrogen \( P_{\text{N}_2} \) is initially zero (since nitrogen is not present in the starting conditions). Therefore, at the start:

\[
Q = 0
\]

Since \( Q < K \), the reaction will proceed forward to increase \( Q \), meaning more nitrogen (\( \text{N}_2 \)) and hydrogen (\( \text{H}_2 \)) will be produced, causing the pressure of \( \text{H}_2 \) to **rise**.

### Step 3: Can adding nitrogen (\( \text{N}_2 \)) reverse the tendency?

To answer this, let’s consider Le Chatelier's principle. Adding nitrogen increases \( P_{\text{N}_2} \), which would increase \( Q \). If \( Q \) exceeds \( K \), the reaction would shift in the reverse direction, causing \( P_{\text{H}_2} \) to decrease. So yes, adding nitrogen can reverse the tendency of \( P_{\text{H}_2} \) to rise.

### Step 4: Calculate the minimum pressure of \( \text{N}_2 \)

To reverse the tendency, we need \( Q \geq K \). Thus:

\[
Q = K = 0.0011
\]

We need to calculate the minimum \( P_{\text{N}_2} \) such that:

\[
Q = \frac{P_{\text{N}_2} P_{\text{H}_2}^3}{P_{\text{NH}_3}} \geq K
\]

Substituting the known values:

\[
0.0011 = \frac{P_{\text{N}_2} (0.628)^3}{4.48}
\]

\[
0.0011 \times 4.48 = P_{\text{N}_2} \times (0.248)
\]

\[
0.004928 = P_{\text{N}_2} \times 0.248
\]

\[
P_{\text{N}_2} = \frac{0.004928}{0.248} \approx 0.020 \, \text{atm}
\]

Thus, the minimum pressure of \( \text{N}_2 \) needed to reverse the tendency is **0.020 atm**.

---

### Summary of Answers:
1. The pressure of \( \text{H}_2 \) will tend to **rise**.
2. Yes, adding nitrogen can reverse the tendency.
3. The minimum pressure of \( \text{N}_2 \) needed to reverse the tendency is **0.020 atm**.

Would you like more details on any part of this? Here are five related questions to consider:

1. How does the equilibrium constant change with temperature?
2. What is the significance of the reaction quotient \( Q \) compared to \( K \)?
3. How would the introduction of a catalyst affect this equilibrium?
4. How does the change in \( \Delta G^0 \) affect the equilibrium constant?
5. What would happen if more ammonia (\( \text{NH}_3 \)) were added to the system?

**Tip**: Always check the reaction quotient to predict which direction a reaction will shift when starting conditions deviate from equilibrium.

Consider the following equilibrium:

2 NH3(g) ↔ N2(g) + 3 H2(g), ΔG0 = 34 kJ.

Now suppose a reaction vessel is filled with 4.48 atm of ammonia NH3 and 0.628 atm of hydrogen H2 at 333°C. Answer the following questions:

1. Under these conditions, will the pressure of H2 tend to rise or fall?
2. Is it possible to reverse this tendency by adding N2?
3. If reversible, calculate the minimum pressure of N2 needed to reverse it. Round your answer to 2 significant digits (atm).

Explore equilibrium concepts and calculate whether the pressure of H2 will rise or fall in a system with 4.48 atm of NH3 and 0.628 atm of H2 at 333°C. Learn to determine if adding N2 can reverse this tendency and find the minimum pressure required for reversal. Step-by-step calculations included.

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