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

Step-by-step explanation:

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

Half life
`(t_(1/2)) = (0.693)/(k)`

Hence, calculate the rate constant as follows.

`t_(1/2) = (0.693)/(k)`

k =
`(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 =
`(0.693)/(13)`

= 0.0533
`h^(-1)`

Hence, calculate the activation energy as follows.

`ln((k_(1))/(k_(2))) = (-E_(a))/(R) * ((1)/(T_(1)) - (1)/(T_(2)))`

`ln((0.00495)/(0.0533)) = (-E_(a))/(8.314) * ((1)/(348) - (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((k_(1))/(k_(2))) = (-E_(a))/(R) * ((1)/(T_(1)) - (1)/(T_(2)))`

`ln((0.00495)/(k_(2))) = (-102600)/(8.314) * ((1)/(348) - (1)/(293))`

`k_(2) = 6.36 * 10^(-6) h^(-1)`

Calculate the time as follows.

kt =
`ln((C_(initial))/(C_(final)))`

`6.36 * 10^(-6) h^(-1) * t = ln ((100)/(50))`

`6.36 * 10^(-6) h^(-1) * t = 0.693`

t = 108985 hours

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