Question

The inversion of cane sugar proceeds with the half-life of 600 minutes at pH=5 for any concentration of sugar. However at pH=6, the half life changes to 60 minute. The rate law expression for sugar inversion can be written as: (if it is first order with respect to sugar):
A.$r = k{\left[ {sugar} \right]^2}{\left[ {{H^ + }} \right]^0}$
B.$r = k{\left[ {sugar} \right]^1}{\left[ {{H^ + }} \right]^0}$
C.$r = k{\left[ {sugar} \right]^2}{\left[ {{H^ + }} \right]^1}$
D.$r = k{\left[ {sugar} \right]^1}{\left[ {{H^ + }} \right]^{ - 1}}$

Answer 1

Hint:We have to know that the first order reaction is the reaction that goes through at a rate that is dependent straightly on concentration of one reactant. The half-life of reaction in which the initial population is reduced by half of its original value and half-life of a first order reactant is constant.

Complete step by step answer:
We know that order of the chemical reaction is the sum of the power of concentration of reactants in the rate law expression.
For a reaction, $xA + yB\xrightarrow{{}}P$
Rate=$k{\left[ A \right]^x}{\left[ B \right]^y}$
By summing up the power of the concentration of reactants we give the order of the reaction.
Order=\[x + y\]
When the pH of the solution is raised by one unit from five to six the concentration of the hydrogen ion decreases to one-tenth. Therefore, the half-life period also changes to one-tenth. The reaction rate increases ten times and the order of the reaction with respect to the concentration of the hydrogen ion ${\text{ - 1}}$. Therefore, the formula for the rate of the reaction is written as,
\[r = k{\left[ {sugar} \right]^1}{\left[ H \right]^{ - 1}}\]
Therefore, the option (D) is correct.

Note:
We have to remember that the rate constant of the forward reaction divided by the rate constant of the reverse reaction gives the equilibrium constant of a reaction. The rate constant of a first order reaction contains time as its unit and not concentration unit. From this we could relate that for a first order reaction, the numerical value of k is independent of the unit that expresses the concentration. Even if we change the concentration unit, the numerical value of k for a first order reaction would not change.