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Chemistry Question on Chemical Kinetics

The inversion of cane sugar is first order in [sugar] and proceeds with half-life of 600 minutes at pH = 4 for a given concentration of sugar. However, if pH = 5, the half-life changes to 60 minutes. The rate law expression for the sugar inversion can be written as

A

rate =k[sugar]1[H+]2=k\left[sugar\right]^{1}\left[H^{+}\right]^{2}

B

rate =k[sugar]1[H+]1=k\left[sugar\right]^{1}\left[H^{+}\right]^{1}

C

rate =k[sugar]1[H+]4=k\left[sugar\right]^{1}\left[H^{+}\right]^{4}

D

rate =k[sugar]1[H+]0=k\left[sugar\right]^{1}\left[H^{+}\right]^{0}

Answer

rate =k[sugar]1[H+]0=k\left[sugar\right]^{1}\left[H^{+}\right]^{0}

Explanation

Solution

Answer (d) rate =k[sugar]1[H+]0=k\left[sugar\right]^{1}\left[H^{+}\right]^{0}