Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5

Q.

For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, ΔCH= -601.70 kJ mol-1, the magnitude of change in internal energy for the reaction is __________ kJ. (Nearest integer)
(Given : R = 8.3 J K-1 mol-1)

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

answer is 600.

(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

detailed_solution_thumbnail

Mg(s)+12O2(g)MgO(s)ΔH=ΔU+ΔngRT601.70×103=ΔU12×8.3×300601.70kJ=ΔU1.245kJΔU=600.455kJ

Watch 3-min video & get full concept clarity

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!

whats app icon