ENTROPY AND FREE ENERGY Using reaction free energy to predict equilibrium composition 0.35 Consider the following equilibrium: N, ()+3H, 6) – 2NH, () Now suppose a reaction vessel is filled with 4.40 atm of nitrogen (N2) and 9.54 atm of ammonia (NH) at 733. °C. Answer the following questions about this AGO =-34, KJ system: Under these conditions, will the pressure of N, tend to rise or fall? rise fall 0.0 ? 5 ? yes no Is it possible to reverse this tendency by adding H,? In other words, if you said the pressure of N, will tend to rise, can that be changed to a tendency to fall by adding HSimilarly, if you said the pressure of N, will tend to fall, can that be changed to a tendency to rise by adding H7 if you said the tendency can be reversed in the second question, calculate the minimum pressure of Hy needed to reverse it. Round your answer to 2 significant digits. 0 | sim
The correct answer and explanation is:
To determine whether the pressure of N2N_2 will tend to rise or fall under the given conditions, we need to analyze the reaction: N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
Step 1: Reaction Quotient (Q) vs. Equilibrium Constant (K)
The reaction’s standard Gibbs free energy change (ΔG∘\Delta G^\circ) is given as −34-34 kJ. We use this to determine the equilibrium constant: ΔG∘=−RTlnK\Delta G^\circ = -RT \ln K
Using R=8.314R = 8.314 J/mol·K and T=733+273=1006T = 733 + 273 = 1006 K, K=e−ΔG∘/RT=e(34000)/(8.314×1006)K = e^{-\Delta G^\circ / RT} = e^{(34000) / (8.314 \times 1006)} K=e4.07≈58.6K = e^{4.07} \approx 58.6
Now, we calculate the reaction quotient QQ: Q=PNH32PN2PH23Q = \frac{P_{NH_3}^2}{P_{N_2} P_{H_2}^3}
Given: PN2=4.40 atm,PNH3=9.54 atm,PH2 initially unknownP_{N_2} = 4.40 \text{ atm}, \quad P_{NH_3} = 9.54 \text{ atm}, \quad P_{H_2} \text{ initially unknown}
Since H2H_2 is not initially mentioned, assume it to be very low, meaning QQ is large.
If Q>KQ > K, the reaction shifts left, increasing N2N_2 pressure. If Q<KQ < K, it shifts right, decreasing N2N_2 pressure.
Because NH3NH_3 is already high and H2H_2 is missing or low, we expect Q>KQ > K, causing a shift left, meaning N2N_2 pressure will rise.
Step 2: Can the Tendency Be Reversed by Adding H2H_2?
Yes. By adding H2H_2, we decrease QQ, pushing the reaction right toward NH₃ formation, which reduces N2N_2.
Step 3: Minimum H2H_2 Pressure Needed
To reverse the trend, we set Q=KQ = K: 9.5424.40×PH23=58.6\frac{9.54^2}{4.40 \times P_{H_2}^3} = 58.6
Solving for PH2P_{H_2}: PH23=9.5424.40×58.6P_{H_2}^3 = \frac{9.54^2}{4.40 \times 58.6} PH23=91.02257.84=0.353P_{H_2}^3 = \frac{91.02}{257.84} = 0.353 PH2=0.3533≈0.71 atmP_{H_2} = \sqrt[3]{0.353} \approx 0.71 \text{ atm}
Conclusion
- The pressure of N2N_2 will rise.
- Adding H2H_2 can reverse the tendency.
- The minimum H2H_2 pressure needed to reverse the tendency is 0.71 atm.
Now, I’ll generate an image representing this equilibrium shift visually.
Here is a visual representation of the equilibrium reaction and how the shift occurs when H2H_2 is added. Let me know if you need any modifications or explanations!
