Q. 194.2( 34 Votes )

For the reaction<

Answer :

∆U = -10.5 kJ

Also, ∆U = ∆H – ∆nRT


Where,


∆U = change in energy


∆H = change in enthalpy


∆n = moles of products – moles of reactants = 2-3 = -1


R = 8.314 × 10-3 kJ mol-1 K-1 - constant


T = temperature; here room temperature is taken as 298 K


Substituting all the above values, we get


∆H = -10.5kJ + (-1 mol × (8.314× 10-3kJ mol-1 K-1) × 298 K


= -12.977kJ


we know that


∆G = ∆H - T∆S, where


∆G = change in Gibbs energy


∆H = change in enthalpy, got as -12.977 kJ


∆S = change in entropy, given as -44.1 JK-1 = -44.1 × 10-3 kJK-1


T = temperature at which reaction occurs = 298K


Substituting the above values


∆G = -12.977 – (298 × -44.1 × 10-3)


= -12.977 – ( -13.14)


= 0.163kJ


As ∆G > 0, reaction doesn’t occur spontaneously. Reactions are spontaneous only when ∆G < 0.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Calculate a) <spaNCERT - Chemistry Part-I