Question

at equilibrium, a reaction vessel contains 4.50 atm of br₂ and 1.10 atm of nbr₃. according to the reaction: 2 nbr₃(g) ⇌ n₂(g) + 3 br₂(g) kₚ = 4.8. determine the equilibrium partial pressure of n₂.

Explanation:

Step1: Write the expression for $K_p$

The expression for the equilibrium - constant $K_p$ for the reaction $2NBr_3(g)
ightleftharpoons N_2(g)+3Br_2(g)$ is $K_p=\frac{P_{N_2}\times P_{Br_2}^3}{P_{NBr_3}^2}$, where $P_{N_2}$, $P_{Br_2}$, and $P_{NBr_3}$ are the equilibrium partial pressures of $N_2$, $Br_2$, and $NBr_3$ respectively.

Step2: Substitute the known values into the $K_p$ expression

We know that $K_p = 4.8$, $P_{Br_2}=4.50$ atm, and $P_{NBr_3}=1.10$ atm. Substituting these values into the $K_p$ expression, we get $4.8=\frac{P_{N_2}\times(4.50)^3}{(1.10)^2}$.

Step3: Solve for $P_{N_2}$

First, calculate $(4.50)^3 = 4.50\times4.50\times4.50=91.125$ and $(1.10)^2 = 1.21$. Then the equation becomes $4.8=\frac{P_{N_2}\times91.125}{1.21}$.
Cross - multiply to get $4.8\times1.21 = P_{N_2}\times91.125$.
$4.8\times1.21 = 5.808$. So, $P_{N_2}=\frac{5.808}{91.125}$.
$P_{N_2}=0.064$ atm.

Answer:

$0.064$ atm