For reactions in the gas phase, an equilibrium constant may be written in terms of molarity (K) or in terms of partial pressures (K,). The value of K, for the reaction shown below is equal to 3.1×10-4 at 298 K. 2NOBR(g) = 2NO(g) + Br, (g) 3rd attempt See Periodic Table O See Hint What is the value for Kp at 298 K for the reaction represented by the equation shown below? The value for the gas constant, R, is 0.08206 L•atm/mol • K. NOB1(g) = NO(g) + Br, (g) 1.7029 x 10 -2
For reactions in the gas phase, an equilibrium constant may be written in terms of molarity (K) or in terms of partial pressures (K,). The value of K, for the reaction shown below is equal to 3.1×10-4 at 298 K. 2NOBR(g) = 2NO(g) + Br, (g) 3rd attempt See Periodic Table O See Hint What is the value for Kp at 298 K for the reaction represented by the equation shown below? The value for the gas constant, R, is 0.08206 L•atm/mol • K. NOB1(g) = NO(g) + Br, (g) 1.7029 x 10 -2
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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![**Equilibrium Constants in Gas-Phase Reactions**
For reactions in the gas phase, an equilibrium constant may be written in terms of molarity (\(K_c\)) or in terms of partial pressures (\(K_p\)).
**Reaction and Known Constant:**
The value of \(K_c\) for the reaction shown below is equal to \(3.1 \times 10^{-4}\) at 298 K.
\[ 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \]
---
### 3rd Attempt
**Objective:**
Determine the value for \(K_p\) at 298 K for the reaction represented by the equation shown below. The value for the gas constant, \(R\), is \(0.08206 \, \text{L} \cdot \text{atm/mol} \cdot \text{K}\).
**Reaction:**
\[ \text{NOBr}(g) \rightleftharpoons \text{NO}(g) + \frac{1}{2} \text{Br}_2(g) \]
**Given Calculation Attempt:**
\[
1.7029 \times 10^{-2}
\]
**Additional Resources:**
- *See Periodic Table*
- *See Hint*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd764935e-b0d6-4dff-8d88-5471e17cdc08%2F267b3206-d9e3-4458-b54d-691e95bf302c%2Fpzkr1e_processed.png&w=3840&q=75)
Transcribed Image Text:**Equilibrium Constants in Gas-Phase Reactions**
For reactions in the gas phase, an equilibrium constant may be written in terms of molarity (\(K_c\)) or in terms of partial pressures (\(K_p\)).
**Reaction and Known Constant:**
The value of \(K_c\) for the reaction shown below is equal to \(3.1 \times 10^{-4}\) at 298 K.
\[ 2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g) \]
---
### 3rd Attempt
**Objective:**
Determine the value for \(K_p\) at 298 K for the reaction represented by the equation shown below. The value for the gas constant, \(R\), is \(0.08206 \, \text{L} \cdot \text{atm/mol} \cdot \text{K}\).
**Reaction:**
\[ \text{NOBr}(g) \rightleftharpoons \text{NO}(g) + \frac{1}{2} \text{Br}_2(g) \]
**Given Calculation Attempt:**
\[
1.7029 \times 10^{-2}
\]
**Additional Resources:**
- *See Periodic Table*
- *See Hint*
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