alde123
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I'm stuck on the question below:

N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H = –92 kJ mol–1

The chemist adds more nitrogen to the equilibrium mixture.The temperature is kept at 300 K and the volume at 5.00 dm3. The chemist predicts that the addition of nitrogen will increase the proportion of H2(g) that reacts.

I answered in terms of both Kc and le Chatelier, but the markscheme says to ignore le Chatelier responses. Why is that?

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Kyle03
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Did you have to calculate Kc beforehand? Also was your explanation correct when describing in terms of Kc
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alde123
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(Original post by Kyle03)
Did you have to calculate Kc beforehand? Also was your explanation correct when describing in terms of Kc
Yes to both, but there was also a question about applying le chatelier's principle before.
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Adam_1999
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Kc will remain the same because the temperature stays constant. Temperature is the only factor that affects Kc. This therefore means the equilibrium position will not change. This is why talking about le chatelier's principle is wrong.

Could you post the full question and mark scheme? You only posted the chemist's prediction.
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alde123
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(Original post by Adam_1999)
Kc will remain the same because the temperature stays constant. Temperature is the only factor that affects Kc. This therefore means the equilibrium position will not change. This is why talking about le chatelier's principle is wrong.

Could you post the full question and mark scheme? You only posted the chemist's prediction.
Other questions:
Ammonia, NH3, is manufactured by the chemical industry from nitrogen and hydrogen gases. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H = –92 kJ mol–1
- An iron catalyst is used which provides several benefits for sustainability.
- The chemical industry uses operational conditions that are different from the conditions predicted to give a maximum equilibrium yield.
(a) Use your understanding of Chemistry to explain the above statements.

(b) A research chemist investigates how the value of Kc changes with temperature. N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H = –92 kJ mol–1
- The chemist mixes 0.800 mol of N2(g) and 2.400 mol of H2(g) and leaves the mixture to reach equilibrium at 300 ºC.
- The total volume of the equilibrium mixture is 5.00 dm3 .
- At equilibrium, 0.360 mol of NH3(g) has formed.

Calculate the value of Kc under these conditions.

Mark-scheme for this question:

IGNORE le Chatelier responses
Each marking point is independent
- Kc Kc does not change (with pressure/ concentration)
- Comparison of conc terms with more N2 [N2] increases OR denominator/bottom of Kc expression increases
- yield of NH3 linked to Kc Chemist is correct AND denominator decreases OR numerator increases to restore equilibrium Kc
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