2.44 m 638cm 0.005 km 700mm smallest to largest

Solved

Show answers

Ответ:

Rainey1664

08.03.2020

Chemistry

The answer to the question is

The enzyme must lower the activation energy by 35624.5 KJ/mol

Explanation:

The relation is

k = $Ae^{\frac{-Ea}{RT} }$

Where k = rate of chemical reaction

A = Arrhenius constant

Ea = Activation energy

T = Temperature

R = Gas constant

at 37 °C we have T = (37 +273.15) K = 310.15 K

k = $Ae^{\frac{-Ea}{(8.314)(310.15)} }$

when a catalyst is introduced we have Ea(catalyst)

kc = $Ae^{\frac{-Eacatalyst}{(8.314)(310.15)} }$

Finding the ratio of the two reaction rate and noting that Kc/K = 1.0 x 10⁶

Thus we have

Kc/k = $\frac{Ae^{\frac{-Eacatalyst}{(8.314)(310.15)} }}{Ae^{\frac{-Ea}{(8.314)(310.15)} }}$ = 1.0 x 10⁶ by taking natural logarithm of both sides, we have

6 × ln (10) = Ln (Kc/k ) = ${\frac{Ea}{(8.314)(310.15)} }- {\frac{Eacatalyst}{(8.314)(310.15)} }$

so that Ea - Eacatalyst = 8.314 × 310.15 × 13.816 = 35624.5 KJ/mol

Therefore the enzyme must lower the activation energy by 35624.5 KJ/molto achieve a million fold increse in the reaction rate constant

0,0(0 оценок)

Ask your question to the AI advisor

Ai-bot is an expert in any field and is the perfect companion for reliable and useful answers and advice on a variety of topics, including science, history, technology, art, sports, health, culture and more.

Ask your question

More tips

Answers on questions: Chemistry

Ask an AI advisor a question

I want advice