In the production of ammonia, gaseous nitrogen is reacted with gaseous hydrogen in the presence a catalyst. The system is allowed to come to equilibrium at 25oC and 101.3 kPa.:N2(g) + 3H2(g) $\ce{ }$ 2NH3(g) $\ce{ $\Delta$ }$H = - 92 kJ mol-1, K = 1.0 x 10-3 M-2A possible numerical value for the equilibrium constant at 50 oC is

Question

In the production of ammonia, gaseous nitrogen is reacted with gaseous hydrogen in the presence a catalyst. The system is allowed to come to equilibrium at 25oC and 101.3 kPa.:N2(g) +  3H2(g) $\ce{ <=> }$ 2NH3(g) $\ce{ $\Delta$ }$H = - 92 kJ mol-1, K = 1.0 x 10-3 M-2A possible numerical value for the equilibrium constant at 50 oC is

BlackTom AI · Accepted Answer

First, identify the reaction and how temperature affects the equilibrium constant. The Haber process reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g) is exothermic (ΔH° = -92 kJ/mol). For an exothermic reaction, increasing temperature shifts the equilibrium toward the reactants and decreases the value of the equilibrium constant K.

We are given K at 25 °C (298 K) as K1 = 1.0 × 10^-3 (with units M^-2 for this reaction). We want K at 50 °C (323 K), call it K2.

Use the van ’t Hoff relation in its integrated form for moderate temperature changes:
ln(K2/K1) = (-ΔH°/R) × (1/T2 − 1/T1)
where ΔH° is taken per mole of reaction as written, and R = 8.314 J·mol^-1·K^-1.

Plug in the values:
- ΔH° = -92,000 J/mol
- T1 = 298 K, T2 = 323 K
- 1/T2 − 1/T1 = (1/323) − (1/298) ≈ 0.003095 − 0.003356 = −0.000261 K^-1

Compute the factor (-ΔH°/R): (-(-92,000 J/mol)) / 8.314 J·mol^-1·K^-1 ≈ 92,000 / 8.314 ≈ 11,070

Now multiply by the temperature difference term:
ln(K2/K1) ≈ 11,070 × (−0.000261) ≈ −2.89

Exponentiate to get the ratio of the equilibrium constants:
K2/K1 ≈ e^(−2.89) ≈ 0.055

Therefore, K2 ≈ K1 × 0.055 ≈ (1.0 × 10^−3) × 0.055 ≈ 5.5 × 10^−5

Interpreting the result, the equilibrium constant decreases with rising temperature for this exothermic reaction, which is consistent with Le Châtelier’s principle. The numerical value is on the order of 10^−5 to a few ×10^−5, i.e., significantly smaller than the 25 °C value.

When comparing to the provided option, note that the exact numerical value depends on rounding during intermediate steps; the estimate places K2 well below 1 × 10^−4, consistent with the general trend for this system. If the options include a value around 1 × 10^−4, that would be somewhat higher than the computed estimate but may reflect rounding choices or simplified calculations.

In summary, because the reaction is exothermic, increasing the temperature from 25 °C to 50 °C decreases K, yielding an order-of-magnitude decrease in K, with a ballpark value near a few ×10^−5. The reasoning aligns with the expectation that K falls at higher temperature for an exothermic formation of ammonia.

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