Question

The hydrolysis of pyrophosphate can be described by the reaction H2PO2 7 +H2O ! 2H2PO 4 If the pH is held constant, the reaction is pseudo-first order with respect to H2PO2– 7 . 126 (a) If, at a given pH, the half-life of pyrophosphate is 140 h at 75C and 13 h at 100C, what is EAr for the reaction? (b) Estimate the time required for 50% hydrolysis of pyrophosphate in a solution at the same pH as in part (a), but at 20C.

Answer 1

Explanation:

Relation between half-life and rate constant is as follows.

       Half life $(t_{1/2}) = \frac{0.693}{k}$

Hence, calculate the rate constant as follows.

           $t_{1/2} = \frac{0.693}{k}$

               k = $\frac{0.693}{140 hr}$

                   = 0.00495 $h^{-1}$

Hence, rate constant at $75^{o}C$ is 0.00495 $h^{-1}$.

Now, at $100^{o}C$, value of rate constant will be as follows.

            k = $\frac{0.693}{13}$

              = 0.0533 $h^{-1}$

Hence, calculate the activation energy as follows.

                 $ln(\frac{k_{1}}{k_{2}}) = \frac{-E_{a}}{R} \times (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

          $ln(\frac{0.00495}{0.0533}) = \frac{-E_{a}}{8.314} \times (\frac{1}{348} - \frac{1}{373})$

            $E_{a}$ = 102.6 kJ/mol

Therefore, value of activation energy is  102.6 kJ/mol .

Now, k at $20^{o}C$,

        $ln(\frac{k_{1}}{k_{2}}) = \frac{-E_{a}}{R} \times (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

         $ln(\frac{0.00495}{k_{2}}) = \frac{-102600}{8.314} \times (\frac{1}{348} - \frac{1}{293})$

         $k_{2} = 6.36 \times 10^{-6} h^{-1}$

Calculate the time as follows.

           kt = $ln(\frac{C_{initial}}{C_{final}})$

        $6.36 \times 10^{-6} h^{-1} \times t = ln (\frac{100}{50})$

       $6.36 \times 10^{-6} h^{-1} \times t = 0.693$

                   t = 108985 hours

Thus, we can conclude that time required for 50% hydrolysis of pyrophosphate is 108985 hours.