The Kp value for the reaction H2(g)+I2(g)⇌2HI(g) at 460oC is 49. If the initial pressure of H2 and I2 are 0.5 atm respectively, determine the partial pressure of H2,I2 and HI gases at equilibrium.
A
0.11 atm, 0.11 atm, 0.77 atm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
0.11 atm, 0.77 atm, 0.11 atm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.77 atm, 0.11 atm, 0.11 atm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 0.11 atm, 0.11 atm, 0.77 atm
H2+I2→2HI
Initial : 0.50.50
At eq. : (0.5−x)(0.5−x)2x
Kp=(2x)2(0.5−x)2=49
On solving the above equation, 2x=3.5−7x
9x=3.5
x=0.389
At equillibrium,Partial pressure of hydrogen, p′H2=0.5−0.389=0.111atm
Partial pressure of I2,p′I2=0.5−0.389=0.111atm
Partial pressure of hydrogen iodide , p′HI=2×0.389=0.77atm