Equilibrium

[IL Pre Med]

This reaction is given:

N2(g) + 3 H2(g) <--> 2 NH3(g)

&#916;H°rxn = –91.8 kJ/mol

In order to maintain the maximum yield of ammonia when raising the temperature of the reaction, scientists should:

A. maintain N2 and H2 at high pressure.
B. allow N2 and H2 to expand with the increased kinetic energy.
C. increase the volume in the reaction chamber to favor the forward reaction.
D. continuously remove the excess heat produced by the forward reaction.

Correct answer is A.

My question is if heat is on the right with ammonia since the forward rxn is exothermic, if you remove heat won't the system shift to right and produce more ammonia?

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[ashtonjam]

Welp I just got that wrong in my head by answering D. It sounded tempting because heat is a product and removing it should drive the reaction right.

Maybe it's more like this: dG = dH - TdS

Enthalpy is a static value and it's negative. -TdeltaS can change.

As it stands, you are increasing the temperature of the reaction (from question stem info), so -TdeltaS is becoming a lot larger in magnitude. However, entropy is negative as the reaction is written; 4 moles of gas becomes 2 moles of gas.

This means as temperature increases, -TdeltaS is an increasingly positive value, which overwhelms the negative fixed value of enthalpy by a long shot. The Haber process uses really high temperature, after all. The free energy (delta G) of reaction is probably positive and not favorable.

Removing just the generated heat from the reaction, in that case, wouldn't make a huge difference since the ambient temperature of the reaction conditions is still super high.

To get around this, you would probably have to rely on squishing ammonia and hydrogen together at high pressures since that's what Le Chatelier's predicts. Maybe this has something to do with changing the value of delta S?

Could be wrong though, let's wait for other people to answer as well.