Doc Brown's Advanced A Level Chemistry Advanced A Level Chemistry - Kinetics-Rates revision notes Part 6

7.4 The hydrogen, iodine and hydrogen iodide equilibrium

H_2(g) + I_2(g) 2HI_(g)  

Advanced A Level Chemistry Kinetics Index

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Case study 4.4 The H₂/I₂/HI equilibrium

Explanation and derivation of orders of reactants and how to write the rate expression

The relationship between rate expressions and K_c equilibrium expressions

• CAN YOU EVER RELATE EQUILIBRIUM EXPRESSIONS WITH KINETIC RATE EXPRESSIONS?

  • AND SHOULD THEY BE RELATED?

• The gaseous phase equilibrium and kinetics involving hydrogen, iodine and hydrogen iodide has been very well studied quantitatively at temperatures of 250–500^oC.

• The reaction is: H_2(g) + I_2(g) 2HI_(g)  

• The reaction mechanism, in either direction, is controlled by an initial bimolecular collision (rds) with a 'transition state' or 'activated complex' consisting of two hydrogen atoms and two iodine atoms.

• The structure of the 'transition state' is not known and there are two possible mechanisms of either 2 or 4 steps.

• However, the proposed mechanism of ...

  • (i) an initial rate determining step (rds) of I₂ + H₂ ==> intermediate state ==> products for the forward reaction,

  • and HI + HI ==> intermediate state ==> products for the backward reaction

  • ... is supported by the kinetics data which shows that ...

• the rate expression for the forward reaction at equilibrium is:

  • rate_f = k_f[H_2(g][I_2(g)]

  • 1st order with respect to both reactants, hydrogen and iodine, 2nd order overall,

• and the rate expression for the backward reaction at equilibrium is:

  • rate_b = k_b[HI_(g]² 

  • 2nd order with respect to the only 'reactant', hydrogen iodide.

• Now the equilibrium expression for the reaction is ...

  • K_c =  [HI_(g]²/[H_2(g][I_2(g)], the equilibrium constant K_c has no units (dimensionless),

  • but since the rate expressions involve the same concentration expressions as the equilibrium expression and the rates of the forward and backward reaction are the same at equilibrium,

    • we can rearrange the rate expressions so that (where f = forward, b = backward)

  • Since for a dynamic equilibrium rate_f = rate_b

  • therefore ...

  • k_f[H_2(g][I_2(g)] = k_b[HI_(g]² 

  • k_f / k_b = [HI_(g]² / [H_2(g][I_2(g)] = a constant at constant temperature

  • and this constant is the equilibrium constant K_c

  • Hence, an equilibrium can be derived from well proven rate expressions.

  • You can write the logic down in another way e.g.

  • [HI_(g]² = rate_b/k_b and [H_2(g][I_2(g)] = rate_f/k_f

  • therefore we can write: K_c = (rate_b/k_b)/(rate_f/k_f) = k_f/k_b (since the 'rates' cancel out)

  • so the equilibrium constant is the ratio of the two rate constants for the forward and backward reactions.

• This a nice simple example to combine the concept areas of equilibrium and rates of reaction, but many other equilibrium reactions are not so simple to analyse in terms of rate expressions!

• When a system is a dynamic equilibrium the rate of the forward reaction = rate of the backward reaction, so here the H₂/I₂/HI concentrations remain constant, but two reactions are simultaneously occurring.

• Four points should be emphasised ...

  1. Rate expressions can only be obtained from experimental results.

  2. If both the rate expressions are known for a true dynamic equilibrium reaction, then it is possible to derive the correct K_c equilibrium expression and the K_c value at a given temperature.

  3. It is NOT possible to derive rate expressions from either (i) the stoichiometric (balanced) equation or (ii) the K_c equilibrium expression.

  4. It is of course possible, to derive the equilibrium expression from the stoichiometric equation, which can of course be verified by experiment, and more importantly, used to predict equilibrium concentrations for a given set of conditions.

• (see also chemical equilibrium)

• Activation energies for the decomposition of hydrogen iodide.

  • 2HI_(g)   H_2(g)  +  I_2(g)

  • Activation energies: (i) uncatalysed 183 kJmol^-1, (ii) Au catalysed 105 kJmol^-1, (iii) Pt catalysed 58 kJmol^-1

  • The surface of both transition metals act as an efficient catalyst.

  • The activation energy for the uncatalysed reverse reaction i.e. formation of hydrogen iodide is 157 kJmol^-1, does this reflect the lowest bond energy of iodine?

  • Bond enthalpies: H-H 436 kJmol^-1, H-I 299 kJmol^-1, I-I 151 kJmol^-1.

  • One proposed mechanism involves a four atom ['activated complex'] produced on collision of two molecules of hydrogen iodide (forward reaction) OR a hydrogen and iodine molecule (backward reaction).

  • 2HI_(g)   [I...H...H...I]    H_2(g)  +  I_2(g)

  • Since this reversible complex formation is a single bimolecular step, it will result in the observed 2nd order rate equations

    • rate forward = k [HI]²   and  rate backward = k [H₂][I₂]

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