Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Book Online Demo
Check Your IQ
Try Test

Courses

Dropper NEET CourseDropper JEE CourseClass - 12 NEET CourseClass - 12 JEE CourseClass - 11 NEET CourseClass - 11 JEE CourseClass - 10 Foundation NEET CourseClass - 10 Foundation JEE CourseClass - 10 CBSE CourseClass - 9 Foundation NEET CourseClass - 9 Foundation JEE CourseClass -9 CBSE CourseClass - 8 CBSE CourseClass - 7 CBSE CourseClass - 6 CBSE Course
Offline Centres

Q.

Calculate ΔH when 2 moles of solid benzoic acid undergo complete combustion at 300K if C6H5COOH(s)+152O2(g)⟶7CO2(g)+3H2O(l) ΔUreaction =−750kJ/mole

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

- 1502.494 kJ

b

- 751.247 kJ

c

- 752.494 kJ

d

- 1501.24 7 kJ

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Enthalpy change for one mole,

ΔH=−750 kJ+−12×8.314×3001000=−750kJ−1.247 kJ=−751.247 kJ

Thus, the enthalpy change for 2 mole,

2×(−751.247)=−1502.497 kJ

Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
whats app icon