## Tuesday, November 4, 2014

### Increasing the Rate of Dissociation by Adding Acid

Sucrose is a disaccharide produced by the hydrolysis of glucose and fructose molecules in which an hydroxyl group (-OH) in each sugar molecule join producing an oxygen bridge between them by releasing an H2O water molecule. One can see the oxygen atom linking the hexagonal ring (glucose) to the pentagonal ring (fructose) in the .gif below.

animation credit: Wikipedia article Sucrose

Dissociating a disaccharide molecule requires a hydration reaction to replace the missing atoms of the simple sugars. There are always hydrogen and hydroxyl radicals normally present in a sugar solution which can react with the sucrose molecules but adding an acid as Wilhelmy did acts as a catalyst to increase the reaction rate.

To study the effect of adding the acid to the solution we need to modify the equation for the reaction and deduce the corresponding reaction rate equation. In the reaction below we can let A correspond to the sucrose molecule, B to the hydrogen molecules present in the solution resulting from the addition of acid and C and D to the glucose and fructose products. We can also let X represent the progress of the reaction in order to determine the amount of sucrose and other molecules present at any given time. The acid molecules easily dissociate so the amount of hydrogen present is essentially equal to the concentration of acid. Plugging the modified amounts into the reaction rate formula again reduces it to a first order differential equation involving the sucrose concentration [A].

The reaction again follows a binomial rate law and we can simplify it by "completing the square" and rewriting the equation using the auxiliary parameters u and v.

The result is a similar formula involving the hyperbolic tangent function but here λ is replaced by βc so the net effect is that the reaction rate constants are multiplied by the factor β increasing the reaction rate considerably. We can check our math by working an example. From the positive polynomial coefficients a, b and c in the rate equation we can compute the constant parameters α and β and κ. Here κ is complex but that isn't a problem since we can use tan(iθ) = i tanh(θ) with a complex θ. The complex κ is easily handled by Mathcad. The complex value is the results from ([A]0 + α)/β being greater than 1.

The results from an iterative evaluation of the reaction rate formula and those from the derived formula produce identical curves for the reaction, justifying the use of the derived formula. Setting the reaction rate polynomial to zero allows us to compute the equilibrium sucrose concentration Aeq which agrees with the result of the numerical computation.

Supplemental (Nov 4): Try k0=1×10-4, k1=1.234567901234567×10-6

Suppemental (Nov 5): For the hyperbolic tangent of a complex number one can use the following identity which is similar to the identity for tanh(u+v).

Even though κ is complex, tanh(κ) is real and consequentially, A is real.

Note also that since tanh(κ + βct) > 0, the new formula for A subtracts 1 from the tanh instead of subtracting the tanh from 1.