SATHEE: Chemical Thermodynamics Question 403

Chemical Thermodynamics Question 403

Question: What is the equilibrium constant if ATP hydrolysis by water produce standard free energy of $- 50 k J / m o l$ under normal body conditions-

Options:

A) $2.66 \times 10^{8}$

B) $5.81 \times 10^{8}$

C) $1.18 \times 10^{7}$

D) $1.98 \times 10^{8}$

Show Answer

Answer:

Correct Answer: A

Solution:

$Δ G = - R T$ ln $K_{e q}$ : Normal body temperature $= 37^{\circ} C$

$\Rightarrow - 50 \frac{K J}{m o l} = 8.314 \frac{J}{K m o l} \times 310, \ln K_{e q}$

$\Rightarrow 19.39 = \ln K_{e q}$

$\Rightarrow K_{e q} = 2.6 \times 10^{8}$

Privacy Policy

Powered by Prutor@IITK

sathee@iitk.ac.in

Media coverage of SATHEE

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Chemical Thermodynamics Question 403

Question: What is the equilibrium constant if ATP hydrolysis by water produce standard free energy of −50kJ/mol under normal body conditions-

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

Question: What is the equilibrium constant if ATP hydrolysis by water produce standard free energy of $- 50 k J / m o l$ under normal body conditions-