Photo AI
For the reaction $$\frac{1}{3}N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3}NH_3(g)$$, the magnitude of K<sub>c</sub> at 25 °C is: A - VCE - SSCE Chemistry - Question 14 - 2020 - Paper 1
Question 14
For the reaction $$\frac{1}{3}N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3}NH_3(g)$$, the magnitude of K<sub>c</sub> at 25 °C is: A. 9 and ΔH = –30.8 kJ mol<sup>–... show full transcript
Worked Solution & Example Answer:For the reaction $$\frac{1}{3}N_2(g) + H_2(g) \rightleftharpoons \frac{2}{3}NH_3(g)$$, the magnitude of K<sub>c</sub> at 25 °C is: A - VCE - SSCE Chemistry - Question 14 - 2020 - Paper 1
Step 1
Calculate K<sub>c</sub> at 25 °C
Answer
For the given reaction, the equilibrium constant K<sub>c</sub> can be determined from the provided options. The values suggest that the reaction favors product formation at equilibrium. Thus, we can eliminate the options with lower K<sub>c</sub> values.
Among the choices:
- A presents K<sub>c</sub> = 9, which may not indicate a strong product favoring.
- B has K<sub>c</sub> = 213, indicating significant product formation.
- C states K<sub>c</sub> = 640, suggesting a strong product favoring.
- D also shows K<sub>c</sub> = 640, but with a different ΔH.
Considering that lower ΔH values would indicate exothermic reactions and align more with the larger K<sub>c</sub>, both B and C remain relevant.
Therefore, upon review, the correct answer is B: 213 and ΔH = –30.8 kJ mol<sup>–1</sup>, as it aligns with expected equilibrium conditions.