I believe the correct answer is d. thermosphere.
Work done = force × distance travelled in the direction of the force
Force= 2(9.81) = 19.62N
WD= 19.62×130= 2550.6Nm
The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
<u>Explanation:</u>
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
