At 2590 °C the equilibrium constant for the reaction: 2 BrCl(g) Br₂(g) + Cl₂(9) is Kp = 0.836. If the initial pressure of BrCl is 0.00672 atm, what are the equilibrium partial pressures of BrCl, Br₂, and Cl₂? p(BrCl) = = p(Br₂) = P(Cl₂)=
At 2590 °C the equilibrium constant for the reaction: 2 BrCl(g) Br₂(g) + Cl₂(9) is Kp = 0.836. If the initial pressure of BrCl is 0.00672 atm, what are the equilibrium partial pressures of BrCl, Br₂, and Cl₂? p(BrCl) = = p(Br₂) = P(Cl₂)=
Chemistry
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Chapter1: Chemical Foundations
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![At 2590 °C, the equilibrium constant for the reaction:
\[ 2 \text{BrCl} (g) \rightleftharpoons \text{Br}_2 (g) + \text{Cl}_2 (g) \]
is \( K_p = 0.836 \). If the initial pressure of BrCl is 0.00672 atm, what are the equilibrium partial pressures of BrCl, Br\(_2\), and Cl\(_2\)?
\[ p(\text{BrCl}) = \underline{\hspace{1cm}}. \]
\[ p(\text{Br}_2) = \underline{\hspace{1cm}}. \]
\[ p(\text{Cl}_2) = \underline{\hspace{1cm}}. \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86624ab7-da2a-4a5b-91b8-0bf4ad8ab366%2F86bad5bc-bd8e-4094-9aba-fef480c22060%2Fj5q4pmn_processed.png&w=3840&q=75)
Transcribed Image Text:At 2590 °C, the equilibrium constant for the reaction:
\[ 2 \text{BrCl} (g) \rightleftharpoons \text{Br}_2 (g) + \text{Cl}_2 (g) \]
is \( K_p = 0.836 \). If the initial pressure of BrCl is 0.00672 atm, what are the equilibrium partial pressures of BrCl, Br\(_2\), and Cl\(_2\)?
\[ p(\text{BrCl}) = \underline{\hspace{1cm}}. \]
\[ p(\text{Br}_2) = \underline{\hspace{1cm}}. \]
\[ p(\text{Cl}_2) = \underline{\hspace{1cm}}. \]
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