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

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