*Xog pəpIAojd For reactions carried out under standard-state conditions, the equation AG=AH- TAS becomes AG" = H - TAS". Assuming AH" and AS are independent of temperature, one can derive the equation: K2 AH T2-T1. In %3D K1 where K1 and K2 are the equilibrium constants at T, and T2, respectively. Given that at 25.0°C, K̟ is 4.63 x 103 for the reaction N,O,(g) 5 2NO,(g) AH° =58.0 kJ/mol calculate the equilibrium constant at 40.0°C. K =
*Xog pəpIAojd For reactions carried out under standard-state conditions, the equation AG=AH- TAS becomes AG" = H - TAS". Assuming AH" and AS are independent of temperature, one can derive the equation: K2 AH T2-T1. In %3D K1 where K1 and K2 are the equilibrium constants at T, and T2, respectively. Given that at 25.0°C, K̟ is 4.63 x 103 for the reaction N,O,(g) 5 2NO,(g) AH° =58.0 kJ/mol calculate the equilibrium constant at 40.0°C. K =
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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![**Chemical Equilibrium and Temperature Dependence**
Enter your answer in the provided box.
For reactions carried out under standard-state conditions, the equation ΔG = ΔH - TΔS becomes:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
Assuming \( \Delta H^\circ \) and \( \Delta S^\circ \) are independent of temperature, one can derive the equation:
\[ \ln \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^\circ}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]
where \( K_1 \) and \( K_2 \) are the equilibrium constants at \( T_1 \) and \( T_2 \), respectively. Given that at 25.0°C, \( K_c \) is \( 4.63 \times 10^{-3} \) for the reaction:
\[ N_2O_4 (g) \leftrightarrow 2NO_2 (g) \]
and \( \Delta H^\circ = 58.0 \text{ kJ/mol} \),
calculate the equilibrium constant at 40.0°C.
**Equation to use:**
\[ K_c = \boxed{} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F239db1bd-bca7-46b2-879b-9072f95b41a3%2F2fff29b2-3a99-4fa8-8227-e4dbfc80c4b6%2F5rudcv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Equilibrium and Temperature Dependence**
Enter your answer in the provided box.
For reactions carried out under standard-state conditions, the equation ΔG = ΔH - TΔS becomes:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
Assuming \( \Delta H^\circ \) and \( \Delta S^\circ \) are independent of temperature, one can derive the equation:
\[ \ln \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^\circ}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]
where \( K_1 \) and \( K_2 \) are the equilibrium constants at \( T_1 \) and \( T_2 \), respectively. Given that at 25.0°C, \( K_c \) is \( 4.63 \times 10^{-3} \) for the reaction:
\[ N_2O_4 (g) \leftrightarrow 2NO_2 (g) \]
and \( \Delta H^\circ = 58.0 \text{ kJ/mol} \),
calculate the equilibrium constant at 40.0°C.
**Equation to use:**
\[ K_c = \boxed{} \]
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