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.