An evacuated reaction vessel is filled with 0.800 atm of N₂ and 1.681 atm of Br₂. When equilibrium is established according to the reaction below, there are 1.425 atm of Br₂ remaining in the vessel. What is the equilibrium partial pressure of NBr₃? 2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g)

the equilibrium partial pressure of NBr₃ is 0.1707 atm

the equilibrium partial pressure of NBr₃ is 0.1707 atm

the equilibrium partial pressure of NBr₃ is 0.1707 atm

Explanation:

2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g)

initial atm 0.80 1.681

Change atm 2x - x - 3x

equlibrium atm 2x 0.80 - x 1.681 - 3x

At equilibrium the Br₂ (g) remaining 1.425atm

Therefore,

1.681atm - 3x = 1.425atm

3x = 0.256

x = 0.0853

Therefore, the equilibrium partial pressure of 2NBr₃ (g) is 2x

= 2 * 0.0853

= 0.1707 atm

the equilibrium partial pressure of NBr₃ is 0.1707 atm