0.20 Reaction equilibrium in the gas phase  (Page 2/8)

Observation 1: reaction equilibrium

We begin by analyzing a significant industrial chemical process, the synthesis of ammonia gas, NH ₃ , from nitrogen and hydrogen:

N ₂ (g) + 3 H ₂ (g) → 2 NH ₃ (g)

If we start with 1 mole of N ₂ and 3 moles of H ₂ , the balanced equation predicts that we will produce 2 moles of NH ₃ . In fact, if we carry out this reaction starting with these quantities of nitrogen and hydrogen at 298 K in a 100.0 L reaction vessel, we observe that the number of moles of NH ₃ produced is 1.91 mol. This “yield” is less than predicted by the balanced equation, but the difference is not due to a limiting reactant factor. Recall that, in stoichiometry, the limiting reactant is the one that is present in less than the ratio of moles given by the balanced equation. In this case, neither N ₂ nor H ₂ is limiting because they are present initially in a 1:3 ratio, exactly matching the stoichiometry. Note also that this seeming deficit in the yield is not due to any experimental error or imperfection, nor is it due to poor measurements or preparation. Rather, the observation is completely reproducible: every measurement of this reaction at this temperature in this volume starting with 1 mole of N ₂ and 3 moles of H ₂ produces 1.91 moles rather than 2 moles. We conclude that the Reaction (2) achieves reaction equilibrium in which all three gases are present in the gas mixture. We can determine the amounts of each gas at equilibrium from the stoichiometry of the reaction. When n(NH ₃ ) = 1.91 mol are created, the number of moles of N ₂ remaining at equilibrium is n(N ₂ ) = 0.045 mol and n(H ₂ ) = 0.135 mol.

It is important to note that we can vary the relative amount of NH ₃ produced by varying the temperature of the reaction, the volume of the vessel in which the reaction occurs, or the relative starting amounts of N ₂ and H ₂ . We shall study and analyze this observation in detail in later sections. For now, though, we demonstrate that the concept of reaction equilibrium is general to all reactions.

Consider the reaction

H ₂ (g) + I ₂ (g) → 2 HI (g)

If we begin with 1.00 mole of H ₂ and 1.00 mole of I ₂ at 500 K in a reaction vessel of fixed volume, we observe that, at equilibrium, n(HI) = 1.72 mol, leaving in the equilibrium mixture n(H ₂ ) = 0.14 mol and n(I ₂ ) = 0.14 mol.

Similarly, consider the decomposition reaction

N ₂ O ₄ (g) → 2 NO ₂ (g)

At 298 K in a 100.0 L reaction flask, 1.00 mole of N ₂ O ₄ partially decomposes to produce, at equilibrium, n(NO ₂ ) = 0.64 mol and n(N ₂ O ₄ ) = 0.68 mol.

Some chemical reactions achieve an equilibrium that appears to be very nearly complete reaction. For example,

H ₂ (g) + Cl ₂ (g) → 2 HCl(g)

If we begin with 1.00 mole of H ₂ and 1.00 mole of Cl ₂ at 298 K in a reaction vessel of fixed volume, we observe that, at equilibrium, n(HCl) is almost exactly 2.00 mol, leaving virtually no H ₂ or Cl ₂ . This does not mean that the reaction has not come to equilibrium. It means instead that, at equilibrium, there are essentially no reactants remaining.

In each of these cases, the amounts of reactants and products present at equilibrium vary as the conditions are varied but are completely reproducible for fixed conditions. Before making further observations that will lead to a quantitative description of the reaction equilibrium, we consider a qualitative model of equilibrium.