Gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm and that of Br2 (g) is 0.493 atm. Find the partial pressures of HBr when the system reaches equilibrium

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

Explanation:

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

answer: strength i guess

explanation:

multiply 2kg by latent heat of vap