The reaction 2NO(g) + Br₂ (g) — 2 NOBr(g) has Kp = 109 at 25°C. If the equilibrium partial pressure of Br2 is 0.0250 atm and the equilibrium partial pressure of NOBr is 0.0857 atm, calculate the partial pressure of NO at equilibrium. Partial pressure = atm
The reaction 2NO(g) + Br₂ (g) — 2 NOBr(g) has Kp = 109 at 25°C. If the equilibrium partial pressure of Br2 is 0.0250 atm and the equilibrium partial pressure of NOBr is 0.0857 atm, calculate the partial pressure of NO at equilibrium. Partial pressure = atm
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![The reaction
\[ 2 \text{NO} (g) + \text{Br}_2 (g) \rightleftharpoons 2 \text{NOBr} (g) \]
has \( K_p = 109 \) at 25°C. If the equilibrium partial pressure of \(\text{Br}_2\) is 0.0250 atm and the equilibrium partial pressure of \(\text{NOBr}\) is 0.0857 atm, calculate the partial pressure of \(\text{NO}\) at equilibrium.
Partial pressure = \_\_\_\_ atm](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16140577-95fc-4112-8b0a-8e4365545066%2F59149b18-4cb9-4636-b268-8f567f670d38%2Feghsywn_processed.png&w=3840&q=75)
Transcribed Image Text:The reaction
\[ 2 \text{NO} (g) + \text{Br}_2 (g) \rightleftharpoons 2 \text{NOBr} (g) \]
has \( K_p = 109 \) at 25°C. If the equilibrium partial pressure of \(\text{Br}_2\) is 0.0250 atm and the equilibrium partial pressure of \(\text{NOBr}\) is 0.0857 atm, calculate the partial pressure of \(\text{NO}\) at equilibrium.
Partial pressure = \_\_\_\_ atm
![An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction:
\[ 3 \text{H}_2(\text{g}) + \text{N}_2(\text{g}) \rightleftharpoons 2 \text{NH}_3(\text{g}) \]
At equilibrium, the concentrations are \([\text{H}_2] = 6.0 \, \text{M}\), \([\text{N}_2] = 12 \, \text{M}\), and \([\text{NH}_3] = 4.0 \, \text{M}\). What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?
\[[\text{H}_2]_0 = \underline{\hspace{2cm}} \, \text{M}\]
\[[\text{N}_2]_0 = \underline{\hspace{2cm}} \, \text{M}\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16140577-95fc-4112-8b0a-8e4365545066%2F59149b18-4cb9-4636-b268-8f567f670d38%2F2clfmy_processed.png&w=3840&q=75)
Transcribed Image Text:An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction:
\[ 3 \text{H}_2(\text{g}) + \text{N}_2(\text{g}) \rightleftharpoons 2 \text{NH}_3(\text{g}) \]
At equilibrium, the concentrations are \([\text{H}_2] = 6.0 \, \text{M}\), \([\text{N}_2] = 12 \, \text{M}\), and \([\text{NH}_3] = 4.0 \, \text{M}\). What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?
\[[\text{H}_2]_0 = \underline{\hspace{2cm}} \, \text{M}\]
\[[\text{N}_2]_0 = \underline{\hspace{2cm}} \, \text{M}\]
Expert Solution
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Step 1: Explaining the proble statement
Answer:
This problem is based on stoichiometric calculation where all the reactant species react in the molar ratio of their stoichiometric coefficients and all the product species formed are also in the molar ratio of their stoichiometric coefficients.
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