Usually we say that if a reaction $$ aA + bB ⇔ cC + dD$$ has ##K_{eq} > 1## the equilibrium is reached with more products than reagents. Otherwise if ##K_{eq} < 1## the inverse reaction is favoured.(adsbygoogle = window.adsbygoogle || []).push({});

Now let's consider the synthesis of ammonia: $$ 1.5H_{2} + 0.5N_{2} = NH_{3} $$

From letterature we know that @ T = 600K ##ΔG_{R} > 1 → K_{eq} < 1 ##. Now let's write the equilibrium equation (assuming perfect gases as equation of state): $$ \frac {P*y_{NH_{3}}} {(P*y_{H_{2}})^{\frac 3 2} * (P*y_{N_{2}})^{\frac 1 2}} = K_{eq} < 1 $$

It seems to me that the fact of ##K_{eq} ## being smaller than 1 is a meaningful information since I can increase the concentration of ammonia (##y_{NH_{3}}##) simply by increasing the Pressure. So if ##P→inf## then ##y_{NH_{3}} → 1## while ## y_{H_{2}} ## and ## y_{N_{2}} → 0 ##.

Conclusion: working at high pressure I have lots of ammonia even thought K_{eq} < 1. Why then should I worry about K_{eq}? I mean if I MUST have products no metter what I just have to work at higher pressure (it this example).

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# What does Keq tells us?

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