The inversion of cane sugar proceeds with half-life of 500 minute at $\mathrm{pH} =5$ for any concentration…

The inversion of cane sugar proceeds with half-life of 500 minute at $\mathrm{pH} =5$ for any concentration of sugar. However if $\mathrm{pH}=6$, the half-life changes to 50 minute. The rate law expression for the sugar inversion can be written as
  1. $\mathrm{r}=\mathrm{K}[\text { sugar }]^{2}[\mathrm{H}]^{6}$
  2. $\mathrm{r}=\mathrm{K}[\text { sugar }]^{1}[\mathrm{H+}]^{0}$
  3. $\mathrm{r}=\mathrm{K}[\text { sugar }]^{0}\left[\mathrm{H}^{+}ight]^{6}$
  4. $\mathrm{r}=\mathrm{K}[\text { sugar }]^{0}\left[\mathrm{H}^{+}ight]^{1}$

Solution

At \(\mathrm{pH}_{5}\), the half-life is independent of the concentration of sugar. Hence, the reaction is of first order in sugar. The rate law expression is rate \(=\mathrm{K}\left[ight.\) sugar \(^{1}\left[\mathrm{H}^{+}ight]^{\mathrm{m}}\). The relationship between half life and \(\left[\mathrm{H}^{+}ight]\)is \(\mathrm{t}_{1 / 2} \alpha\left[\mathrm{H}^{+}ight]^{1-\mathrm{m}}\) \(500 \alpha\left[10^{-5}ight]^{1-\mathrm{m}}\) \(50 \alpha\left[10^{-6}ight]^{1-\mathrm{m}}\) \(10 \alpha(10)^{1-\mathrm{m}}\) \(1-\mathrm{m}=1\) \(\mathrm{m}=0\) Hence, the rate law expression is \(\mathrm{r}=\mathrm{k}[\mathrm{sugar}]\left[\mathrm{H}^{+}ight]^{0}\). ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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