Question

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)

Answer 1

Answer:

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

Answer 2

The equilibrium partial pressure of $NBr_3$ will be "0.171 atm"

Given equation,

• $2NBr_3(g) \rightleftharpoons N_2 (g) +3Br_2 (g)$

At equilibrium,

• $N_2(g) = 0.800 \ atm$
• $3Br_2(g)= 1.681 \ atm$

Now,

The change of partial pressure of $Br_2$ will be:

= $1.681-1.425$

= $0.256 \ atm$

The increase of partial pressure of $NBr_3$ will be:

= $0.256\times \frac{2}{3}$

= $0.171 \ atm$

hence,

If the initial pressure = 0 atm, then The equilibrium partial pressure of $NBr_3$ = 0.171 atm

Thus the above response is right.

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