SKH Thermodynamics Labs

Lab 12 · Chemical reaction equilibrium

How far does a gas reaction go?

A synthetic dimer dissociates in the gas phase, A₂ ⇌ 2A. At 350 K, ΔG° = −4.0 kJ mol⁻¹ and ΔH° = +50 kJ mol⁻¹ with a standard pressure of 1 bar. One mole of A₂ is fed at 350 K and 100 kPa, and the gases are ideal.

Find the extent of reaction and the equilibrium composition. Then predict how pressure, an inert gas and temperature move it.

2027 prototype
Review the student workflow

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Core: predict, calculate, read the Gibbs-energy curve and explain the result. Steps 1–5.

Graduate: the core steps plus a temperature study of K, the driving force of reactor samples and the standard-state convention in Step 4.

Graduate route. Complete Steps 1–3 for the starting case. In Step 4, carry K to other temperatures with different thermal data, evaluate the driving force of gas samples, and test pressure, inert gas and the standard state. Go to the graduate investigation.

Choose a caseAll cases feed 1 mol of A₂.

Equilibrium is the minimum of G along the reaction

One reaction, ideal gases
Gibbs energy along the extent of reaction at two pressures G minus G at zero extent, divided by RT, against the extent of reaction. At 100 kPa the curve has its minimum at extent 0.705. At 1000 kPa the minimum moves to extent 0.300 and is shallower. Both curves start at zero for pure A2. pure A₂ 1000 kPa · minimum at ξ = 0.300100 kPa · minimum at ξ = 0.705 00.5110−1−2 Extent of reaction ξ / mol [G(ξ) − G(0)]/RT

Model-generated at 350 K. Raising the pressure penalizes the extra mole of gas, so the minimum moves to a smaller extent. K is the same for both curves.

The equilibrium condition for A2 to 2A K equals y_A squared over y_A2 times P over P standard. K depends only on temperature through delta G standard. The composition term and the pressure ratio depend on the state. With one mole of A2 fed, the extent is the square root of K over K plus 4 P over P standard. T onlycompositionpressure · Δν = +1 K=yA²/yA₂×P/P° K = exp(−ΔG°/RT) = exp(4000/(8.314 × 350)) = 3.95 1 mol A₂ fed: yA = 2ξ/(1 + ξ), yA₂ = (1 − ξ)/(1 + ξ)ξ = [K/(K + 4P/P°)]1/2 = (3.95/7.95)1/2 = 0.705

Pressure and inert gas change the state term, not K. Temperature changes K.

At fixed T and P, the system moves along the reaction until G is lowest. At that point ΔG = ΔG° + RT ln Q = 0, so Q = K.

1. Set the reactor and the reaction data

A₂ ⇌ 2A · ideal gas
K
kPa
mol
mol
Standard reaction properties at 350 K · P° = 1 bar

Synthetic teaching values. K at other temperatures follows the van ’t Hoff equation with constant ΔCp°.

Optional. Your choice does not change the calculation.

Add a calculated result to the comparison before changing a setting.

Model and validity

One reaction among ideal gases, with activities aᵢ = yᵢP/P° and P° = 100 kPa. The inert gas is added at constant total pressure. The extent is found from ln Q(ξ) = ln K, which has a single root because G is convex along the reaction. Temperatures are limited to 250–650 K and pressures to 5000 kPa.

Open the current Lab 12 for your own reaction table, formation Gibbs energies and measured activities.

2. Read the result

Calculate to see the extent of reaction and the composition.

4. Check your understanding

Write first, then compare

Use Steps 1–3 to answer each question in your own words. Then open the explanation and compare. Your answers are saved with your study.

A. Solve the starting case on paper

Calculate K from ΔG°, write the mole fractions in terms of ξ, and solve for ξ.

Compare with the explanation

K = exp(4000/(8.314 × 350)) = 3.95. With nA₂ = 1 − ξ, nA = 2ξ and ntotal = 1 + ξ, the condition is K = 4ξ²/(1 − ξ²) × P/P°. At P = P°, ξ² = 3.95/7.95, so ξ = 0.705 and yA = 1.41/1.705 = 0.827.

A negative ΔG° means K > 1, not complete reaction. The equilibrium mixture still contains 0.295 mol of A₂.

B. Why do pressure and an inert gas change the result?

Compare the starting case with “Ten times the pressure” and “Add 4 mol inert gas”. Read K in each case.

Compare with the explanation

K stays 3.95 in all three cases. At 1000 kPa, ξ falls to 0.300; with 4 mol inert at 100 kPa, ξ rises to 0.870.

The reaction makes one extra mole of gas (Δν = +1), so Q contains the factor P/P° × 1/ntotal. Higher pressure raises Q, so the reaction shifts back. An inert gas at constant total pressure lowers the partial pressures, so the reaction shifts forward. In a rigid vessel, adding an inert gas would leave the partial pressures and ξ unchanged.

C. What does temperature do to K?

Load “Heated to 400 K” and compare K and ξ. Then calculate 300 K.

Compare with the explanation

ln K(400) = ln K(350) + (ΔH°/R)(1/350 − 1/400) = 1.375 + 2.148 = 3.522, so K = 33.9 and ξ = 0.946. At 300 K, K = 0.226 and ξ = 0.231.

The dissociation is endothermic, so K rises with temperature. Unlike pressure, temperature changes K itself, because ΔG° depends on T.

4. Investigate the equilibrium evidence

Master's core work

Calculate a state in Step 1 before running an investigation.

A. Carry K to other temperatures

  1. Start from ΔG° at 350 K and compute ln K from 300 to 550 K with constant ΔH°, with ΔCp° = +30 J mol⁻¹ K⁻¹, and with ΔH° ± 10%.
  2. Compare ln K and the fraction dissociated at the current pressure.
  3. Explain where the thermal data matter for ln K but not for the conversion, and where they matter for both.

B. Which way will a sample react?

  1. For four gas samples at the current temperature, compute Q, ln(Q/K) and ΔG = RT ln(Q/K).
  2. Decide the direction of reaction for each sample.
  3. Explain why the same composition can react in opposite directions at different pressures.

C. Pressure, inert gas and the standard state

  1. Compare the current state with ten times and one tenth the pressure, and with 4 mol of inert gas added.
  2. Convert K to a standard pressure of 1 atm and use it consistently. Then use the 1 bar K with a 1 atm standard state by mistake.
  3. Explain why the numerical value of K depends on P° while the physical equilibrium does not.

D. Defend a limited conclusion

The record check only confirms that the sections and calculations are present. Scientific reasoning still needs instructor review.

5. Compare and save

  1. Calculate and save the four named cases.
  2. Find a pressure at 350 K that gives the same ξ as adding 4 mol inert gas at 100 kPa.
  3. Answer questions A–C in Step 4 using the saved comparison.
  4. Download the complete study, reopen it, and reproduce a saved case.
  1. Calculate and save the four named cases.
  2. Complete investigations A–C and write the graduate claims.
  3. Download the complete study, reopen it, and reproduce a saved case and its investigation record.

Your calculated cases will appear here after you add them.

Starting case loaded. Press Calculate the equilibrium.

Your inputs, comparisons and explanations are kept in this open page. Download one study file to keep the complete record and reopen it later. Results are recalculated on restore.