QuestionJuly 19, 2025

At equilibrium, a reaction vessel contains 4.50 atm of Br_(2) and 1.10 atm of NBr_(3) According to the reaction: 2NBr_(3)(g)leftharpoons N_(2)(g)+3Br_(2)(g)Kp=4.8 Determine the equilibrium partial pressure of N_(2)

At equilibrium, a reaction vessel contains 4.50 atm of Br_(2) and 1.10 atm of NBr_(3) According to the reaction: 2NBr_(3)(g)leftharpoons N_(2)(g)+3Br_(2)(g)Kp=4.8 Determine the equilibrium partial pressure of N_(2)
At equilibrium, a reaction vessel contains 4.50 atm of Br_(2) and 1.10 atm of NBr_(3) According to the reaction: 2NBr_(3)(g)leftharpoons N_(2)(g)+3Br_(2)(g)Kp=4.8 Determine the equilibrium partial pressure of N_(2)

Solution
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Answer

0.0637 atm Explanation 1. Write the equilibrium expression For the reaction 2NBr_{3}(g) \rightleftharpoons N_{2}(g) + 3Br_{2}(g), the equilibrium constant expression is K_p = \frac{P_{N_2} \cdot P_{Br_2}^3}{P_{NBr_3}^2}. 2. Substitute known values Given P_{Br_2} = 4.50 atm and P_{NBr_3} = 1.10 atm, substitute these into the expression: 4.8 = \frac{P_{N_2} \cdot (4.50)^3}{(1.10)^2}. 3. Solve for P_{N_2} Rearrange to find P_{N_2}: P_{N_2} = \frac{4.8 \cdot (1.10)^2}{(4.50)^3}. 4. Calculate P_{N_2} Perform the calculation: P_{N_2} = \frac{4.8 \cdot 1.21}{91.125} = \frac{5.808}{91.125} \approx 0.0637 atm.

Explanation

1. Write the equilibrium expression<br /> For the reaction $2NBr_{3}(g) \rightleftharpoons N_{2}(g) + 3Br_{2}(g)$, the equilibrium constant expression is $K_p = \frac{P_{N_2} \cdot P_{Br_2}^3}{P_{NBr_3}^2}$.<br /><br />2. Substitute known values<br /> Given $P_{Br_2} = 4.50$ atm and $P_{NBr_3} = 1.10$ atm, substitute these into the expression: $4.8 = \frac{P_{N_2} \cdot (4.50)^3}{(1.10)^2}$.<br /><br />3. Solve for $P_{N_2}$<br /> Rearrange to find $P_{N_2}$: $P_{N_2} = \frac{4.8 \cdot (1.10)^2}{(4.50)^3}$.<br /><br />4. Calculate $P_{N_2}$<br /> Perform the calculation: $P_{N_2} = \frac{4.8 \cdot 1.21}{91.125} = \frac{5.808}{91.125} \approx 0.0637$ atm.
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