Question

a).what is the relationship between the factors of the Arrhenius equation and the factors of the Eyring equation? ( look at the Arrhenius equation which parts of the Eyring equation are they equal to?

b) what is the steady-state approximation, and how is it used? when is it a poor approximation?

Please help me answer these two question!. Thank you

Answer 1

Answer (a) :-

In general, for a bimolecular reaction, we can write the equation in two ways as shown below: 10. AHİRT ASTR However, since k at a given temperature can have only one value, we should be able to derive relationships between A and AS and betweern Ea and AH'. To obtain this relationship between E and AH', we take the natural log of both sides and then differentiate both expressions for K with respect to T, obtaining the results in (2) and (3) 10 ASTR Multiplying through by RT2 on both sides of the various terms in Equation (3), we obtain the desired result, as shown in Equation () Now we proceed to obtain a relationship between A and AS. In this derivation we will make use of Equation (4). We can write the following equation: AH/RT -RT /RT Note that we have substituted Equation (4) for Ea to obtain the second term. The first two terms in (5) can be rewritten as

Now, looking at this equation, elA is equal to the third term in (5) Thus, we can write Multiplying through be e on both sides, we find as our final result the equation given in (8) A - 21010elTeASIR (8) While we have derived this equation for a bimolecular reaction process in solution, it holds for reactions of any molecularity occurring in solution Note that we have to specify a temperature when we make calculations of A and E, using (4) and (8). Usually this is taken to be the average temperature over the range where measurements are made. This is valid if the temperature range is not too large. Thus, these conversions result in A and E, values that are "averages" over this limited temperature range. (In general, we regard AHt and Δβ as constants and E and A as being the quantities that are dependent on temperature.) For gas phase reactions, a different relationship holds. In these cases, E, is related to AFF as shown in (9) In this equation, An' is the change in the number of molecules changed in going from the reactants to the transition state. For example, a bimolecular reaction in the gas phase would have Δ㎡--1 and, thus E AH- 2RT. Correspondingly, a different relationship holds between A and AS# than that given above, with e2 rather than ei appearing in the equation analogous to (8).

Answer (b) :- We apply steady state approximation for reactions where the concentration of the reaction intermediate can be considered to be constant. Steady state means an chemical equilibrium, i.e. the rate of formation equals the rate of consumption.

We need steady state approximation to simplify the derivation of rate laws of many step reactions.

Clearly the steady state approximation is applicable for the case kl 《 k2 where the concen- tration of the intermediate is small and it varies slowly and can be considered to be constamt 0). situation most of the time during the reaction (i.e. to a good approximation, The situation 《 k 2 also means that the intermediate is very reactive and this step is very fast. Also note that the first step is relatively very slow hence the first step determines the rate of the entire reaction. Now let us apply the steady state approximation for this case to derive the rate law In most of the cases, we look for an expression for c (I) in terms of the rate constants and the concentration of reactants. Let us apply t to eq. 5 which is t k1 (10) Using the above equation we can rewrite the rate laws (differential and integral) as follows dot(A) dt -kıcı (A) = de(I) (12) dt det(B) dt kic. (A) (13) = and (A)(A)e c(I)constant (14) (15) k1 -kit (16) where in the last equation we have used 0 because K2 is large

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