Question

An elevated level of the enzyme alkaline phosphatase (ALP) in the serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?

Short answer

The energy of activation for the reaction is 42.99 kJ/mol.

Step-by-step solution

Using Arrhenius Equation

From the Arrhenius Equation, the rate of reaction at two different temperatures is given below

\({\bf{log}}\frac{{{{\bf{k}}_{\bf{2}}}}}{{{{\bf{k}}_{\bf{1}}}}}{\bf{ = }}\frac{{{{\bf{E}}_{\bf{a}}}}}{{{\bf{2}}{\bf{.303}} \times {\bf{R}}}}\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{1}}}}}{\bf{ - }}\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{2}}}}}} \right)\)

where \({{\bf{k}}_{\bf{1}}}\)and \({{\bf{k}}_{\bf{2}}}\)are the rate constants at \({{\bf{T}}_{\bf{1}}}\) and \({{\bf{T}}_{\bf{2}}}\) respectively. Where \({{\bf{T}}_{\bf{1}}}{\bf{ < }}{{\bf{T}}_{\bf{2}}}\). E_ais the activation energy (in J) and R is the gas constant.

From the question, the ratio of \(\frac{{{{\bf{k}}_{\bf{2}}}}}{{{{\bf{k}}_{\bf{1}}}}}{\bf{ = 1}}{\bf{.47}}\)

\(\begin{align}{{\bf{T}}_{\bf{1}}} &= 30^\circ C &= 303K\\{{\bf{T}}_{\bf{1}}} &= 37^\circ C &= 310K\end{align}\)

Calculation of Activation Energy

Replacing the values in the Arrhenius equation,

\(\begin{align}\log 1.47 &= \frac{{{E_a}}}{{2.303 \times 8.314}}\left( {\frac{1}{{303}} - \frac{1}{{310}}} \right)\\{E_a} &= 1.345 \times 2.303 \times 8.314 \times \left( {\frac{{303 \times 310}}{{310 - 303}}} \right)\\{E_a} &= 42988J/mol\\{E_a} &= 42.99kJ/mol\end{align}\)