If Kp = 7.3 xX 10-6 for the reaction below at 400.0 K, then what is the value of Kc?(R = 0.0821L· atm/mol · K.) 2 NB13 (g) = N2 (g) + 3 Br2 (g)

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Problem Statement:**

Given that \( K_p = 7.3 \times 10^{-6} \) for the reaction below at 400.0 K, determine the value of \( K_c \). 

The gas constant \( R = 0.0821 \, \text{L} \cdot \text{atm/mol} \cdot \text{K} \).

**Chemical Reaction:**

\[ 2 \text{NBr}_3 \, (g) \rightleftharpoons \text{N}_2 \, (g) + 3 \text{Br}_2 \, (g) \]

**Concepts Used:**

The relationship between \( K_p \) and \( K_c \) is given by the equation:

\[ K_p = K_c (RT)^{\Delta n} \]

where:
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in moles of gas (moles of products - moles of reactants).
Transcribed Image Text:**Problem Statement:** Given that \( K_p = 7.3 \times 10^{-6} \) for the reaction below at 400.0 K, determine the value of \( K_c \). The gas constant \( R = 0.0821 \, \text{L} \cdot \text{atm/mol} \cdot \text{K} \). **Chemical Reaction:** \[ 2 \text{NBr}_3 \, (g) \rightleftharpoons \text{N}_2 \, (g) + 3 \text{Br}_2 \, (g) \] **Concepts Used:** The relationship between \( K_p \) and \( K_c \) is given by the equation: \[ K_p = K_c (RT)^{\Delta n} \] where: - \( R \) is the gas constant, - \( T \) is the temperature in Kelvin, - \( \Delta n \) is the change in moles of gas (moles of products - moles of reactants).
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