2. Consider the simultaneous formation of multiple products in a chemical reacti

Question

2. Consider the simultaneous formation of multiple products in a chemical reaction. Rate law dt a. Referring to problem lb above, find tf for this process. b. Given the initial concentrations of products (Bl and Cl are both zero, find kinetic expressions for the formation of Bland [Cl as a function of time in terms of LAl c. What is the final value of Cl once the reaction is complete? d. Plot [A], LB] and [Cl as a function of time on the same graph for the case k 4s and k 0.1s e. Show that the observed activation energy Ea for the disappearance of A is given by E.-MEL where Ei is the activation energy for the formation of Band E is the activation energy for the formation of C Hint: start with 2 plug in Arrhenius expressions for k and ka i.e dT RT ki A1 exp(-E1/RT) and k2 A2exp(-E2/Rt), and remember that dT ak dT dT

Explanation / Answer

We are given here,

A -------> B with a rate constant of k1

and,

A --------> C with a rate constant of k2

Thus, Rate of reaction can be represented as ,

Rate = k1[A] ...............From 1st reaction

and

Rate = k2[A]...............From 2nd reaction

The overall reaction Rate will be,

Rate = k1[A] + k2[A]

         = (k1 + k2)[A]

Also represented as the rate of change of [A], that is,

Rate = -d[A]/dt

         = (k1 + k2)[A]

(b) For the kinetic expression let us consider the Rate of the reverse reactions,

Assuming the rate constant for the reverse reaction as k-1 for 1sy reaction and k-2 for 2nd reaction, we can write the Rate equation as,

Rate(reverse) = k-1[B]..............From 1st reaction

and,

Rate(reverse) = k-2[C]..............From 2nd reaction

At equilibrium the Rate of forward reaction equals to the Rate of reverse reaction, thus for the 1st reaction,

Rate = k1[A] = k-1[B]

or,

[B] = k1/k-1[A]

similarly we can write for the 2nd reaction,

Rate = k2[A] = k-2[C]

or,

[C] = k2/k-2[A]

(c) The expression for [B]/[C] once the reaction is complete will be as from (b),

[B]/[C] = (k1/k-1[A]) / (k2/k-2[A])

           = (k1.k-2) / (k-1.k2)

(d) graph is plotted as, put [A] on left hand side y-axis, put [B] and [C] on right hand side y-axis. x-axis is time of reaction.

In the graph the concentration of [A] would reduce over the period of time of reaction. On the other had the concentration of [B] and [C] would be expected to increase over the period of time of reaction.

(e) For activation energy the general equation can be written as,

k = Ae^(-Ea/RT)

or,

ln k = ln A - Ea/RT

Now for 1st and 2nd reaction we can write the activation energy equation as,

ln k1 = ln A1 - Ea1/RT

and,

ln k2 = ln A2 - Ea2/RT

Thus following the overall reaction we can write,

d ln(k1 + k2) / dT = Ea/RT^2

Substituting the k1 and k2 activation energies we get,

d ln(A1 - Ea1/RT + A2 - Ea2/RT) = Ea/RT^2

lets say the energy of activation for k1 stands as E1 and for k2 stands as E2 then,

d ln[E1 + E2] = Ea/RT^2

We know that,

d ln(k1 + k2)/dT = d ln(k1 + k2).dk1/dk1.dT + d ln(k1 + k2).dk2/dk2.dT

substitute with k1 and k2 and energy of activation for each we get,

Ea = k1E1 + k2E2/k1+k2

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