wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The reaction between N2 and H2 to form ammonia has KC=6×102 at the temperature 500oC. The numerical value of Kp for this reaction is;

A
1.5×105
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1.5×105
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.5×106
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.5×106
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 1.5×105
N2+3H22NH3

Given : KC=6.0×102:T=500C=500+273=773K

Δn=231=2

Kp=Kc(RT)Δn=6×102×(0.0821×773)2

=6×102×(63.46)2=6×102(63.46)2=6×1024027.17=0.00148×102

Kp=1.5×102×103=1.5×105

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Enthalpy
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon