# How would an increase in the concentration of a sucrose solution affect the rate of its acidic hydrolysis?

Regarding the acidic hydrolysis of sucrose

$$\ce{\underset{sucrose}{C12H22O11} + H2O + \underset{catalyst}{H+} -> \underset{glucose}{C6H12O6} + \underset{fructose}{C6H12O6} + H+},$$

I think that its mechanism can be described as a 2-step reaction consisting of a step which involves the H+, and then the step which involves the $\ce{H2O}$, and I believe the latter step to be a rate limiting step.

My question in detail is:

Would an increase in the concentration of a sucrose solution, with a constant $\ce{H+}$ concentration, have a 1-to-1 influence on reaction rate (for example, if doubling the concentration of the sucrose solution would result in a hydrolysis reaction twice as fast), or would the addition of $\ce{H2O}$ in the second step significantly limit the change in reaction rate?

• You mention that you believe the second step of the hydrolysis to be rate limiting already. So that would mean that the concentration of sucrose is not. The second step cannot be varied much, assuming water is the solvent (its concentration cannot be changed then) – Phanz Sep 16 '17 at 12:03
• The solvent is water, I should have specified so. – user47268 Sep 16 '17 at 13:05
• I guess my problem has to do more with the concentration dependency of 2-step reactions in general. I an unclear as to how a change in the concentration or quantity of a reactant affects the rate of a 2-step reaction with a rate limiting step. – user47268 Sep 16 '17 at 13:08

In the first step, $H^+$ is consumed. Since it is not rate limiting, I can assume the intermediate $[sucrose.H^+]$ to be in pseudo-equilibrium with $[H^+]$&$[sucrose]$, on the timescale of the second reaction step.
$$r_2 = k_2[sucrose.H^+]$$