QuestionJuly 24, 2025

At high temperatures bromine molecules can dissociate into bromine atoms. For the reaction Br_(2)(g)leftharpoons 2Brcdot (g) K_(p)=2.48times 10^-3 at 1650^circ C A 5.00 L vessel at 1650^circ C is filled with Br_(2)(g) at an initial pressure of 6.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of Brcdot (g) at equilibrium?

At high temperatures bromine molecules can dissociate into bromine atoms. For the reaction Br_(2)(g)leftharpoons 2Brcdot (g) K_(p)=2.48times 10^-3 at 1650^circ C A 5.00 L vessel at 1650^circ C is filled with Br_(2)(g) at an initial pressure of 6.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of Brcdot (g) at equilibrium?
At high temperatures bromine molecules can dissociate into bromine atoms. For the reaction Br_(2)(g)leftharpoons 2Brcdot (g) K_(p)=2.48times 10^-3 at 1650^circ C A 5.00 L vessel at 1650^circ C is filled with Br_(2)(g) at an initial pressure of 6.00 atm and allowed to come to equilibrium. What will be the pressure (in atm) of Brcdot (g) at equilibrium?

Solution
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Answer

The pressure of Br\cdot (g) at equilibrium is approximately 0.34 atm. Explanation 1. Define Initial Conditions Initial pressure of Br_2 is 6.00 atm, and no Br\cdot initially. 2. Set Up Equilibrium Expression For the reaction Br_{2}(g) \rightleftharpoons 2Br\cdot (g), the equilibrium expression is K_p = \frac{(P_{Br})^2}{P_{Br_2}}. 3. Express Pressures in Terms of Change Let x be the change in pressure for Br_2. At equilibrium, P_{Br_2} = 6.00 - x and P_{Br} = 2x. 4. Substitute into Equilibrium Expression Substitute into K_p: 2.48 \times 10^{-3} = \frac{(2x)^2}{6.00 - x}. 5. Solve for x Rearrange to get 4x^2 = 2.48 \times 10^{-3} \times (6.00 - x). Solve this quadratic equation for x. 6. Calculate Pressure of Br\cdot P_{Br} = 2x.

Explanation

1. Define Initial Conditions<br /> Initial pressure of $Br_2$ is 6.00 atm, and no $Br\cdot$ initially.<br /><br />2. Set Up Equilibrium Expression<br /> For the reaction $Br_{2}(g) \rightleftharpoons 2Br\cdot (g)$, the equilibrium expression is $K_p = \frac{(P_{Br})^2}{P_{Br_2}}$.<br /><br />3. Express Pressures in Terms of Change<br /> Let $x$ be the change in pressure for $Br_2$. At equilibrium, $P_{Br_2} = 6.00 - x$ and $P_{Br} = 2x$.<br /><br />4. Substitute into Equilibrium Expression<br /> Substitute into $K_p$: $2.48 \times 10^{-3} = \frac{(2x)^2}{6.00 - x}$.<br /><br />5. Solve for $x$<br /> Rearrange to get $4x^2 = 2.48 \times 10^{-3} \times (6.00 - x)$. Solve this quadratic equation for $x$.<br /><br />6. Calculate Pressure of $Br\cdot$<br /> $P_{Br} = 2x$.
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