H2(g)+I2(g)⇌ 2HI(g), Kp=0 35 at 298 K If PH2=0 10 atm,PI2=0 10 atm andPHI=0 80 atm at 298 K,has equilibrium - Chemistry - Equilibrium

Question

H2(g)+I2(g)⇌ 2HI(g), Kp=0
35 at 298 K
If PH2=0 10
atm,PI2=0 10 atm andPHI=0 80 atm
at 298 K,has equilibrium been reached If not state on which side of
the equilibrium the system is??

veena Vasandani · Accepted Answer

Determine the Qp for the reaction

Qp = p(HI)2/(pH2) (pI2)

pHI = 0 8atm

pH2 = 0 1atm

pI2 = 0 1atm

Qp = (0 8) X (0 8)/(0 1) (0 1) = 64

Kp = 0 35 atm

Qp> Kp

As value of Qp is different from Kp, the reaction is not at
equilibrium

As Qp > Kp, the reaction is on the right side of the equilibrium
and more HI needs to be converted to hydrogen and iodine to reach
equilibrium state

H2(g)+I2(g)⇌ 2HI(g), Kp=0.35 at 298 K If PH2=0.10 atm,PI2=0.10 atm andPHI=0.80 atm at 298 K,has equilibrium been reached.If not state on which side of the equilibrium the system is??

H₂(g)+I₂(g)⇌ 2HI(g), K_p=0.35 at 298 K

If ^PH₂=0.10 atm,^PI₂=0.10 atm and^PHI=0.80 atm at 298 K,has equilibrium been reached.If not state on which side of the equilibrium the system is??