### Calculating \( \Delta G^\circ \) from Equilibrium ConstantsTo determine the standard Gibbs free energy change (\( \Delta G^\circ \)) for a reaction, we can use the equilibrium constant (\( K_p \)) at a given temperature. The table below provides the necessary data to perform these calculations for several reactions:| Reaction | Temperature (°C) | \( K_p \) value ||----------------------------------------------|------------------|------------------|| \( \text{N}_2 (g) + \text{O}_2 (g) \rightarrow 2\text{NO} (g) \) | 2000 | \( 4.1 \times 10^{-4} \) || \( \text{H}_2 (g) + \text{I}_2 (g) \rightarrow 2\text{HI} (g) \) | 400 | 50.0 || \( \text{CO}_2 (g) + \text{H}_2 (g) \rightarrow \text{CO} (g) + \text{H}_2\text{O} (g) \) | 980 | 1.67 || \( \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \) | 900 | 1.04 |### Explanation of Data- **Reaction**: The balanced chemical equation for each reaction.- **Temperature (°C)**: The temperature at which the equilibrium constant \( K_p \) is measured.- **\( K_p \) value**: The equilibrium constant for the reaction at the given temperature.### Calculation of \( \Delta G^\circ \)The relationship between \( \Delta G^\circ \) and \( K_p \) is given by the equation:\[ \Delta G^\circ = -RT \ln (K_p) \]where:- \( R \) is the universal gas constant (8.314 J/(mol·K)).- \( T \) is the temperature in Kelvin (K).#### Steps to Calculate \( \Delta G^\circ \):1. **Convert Temperature to Kelvin**: \[ T(K) = T(°C) + 273.15 \]2. **Calculate \( \Delta

Answer: Relation between equilibrium constant and standard Gibbs free energy change is shown below:…

Calculate AG°for each of the following reactions from the equilibrium constant at the temperature given. Reaction Temperature (°C) N₂ (g) + O₂ (g) → 2NO (g) 2000 H₂(g) + 1₂ (g) →→→ 2HI (g) 400 CO₂ (g) + H₂ (g) → CO (g) + H₂O (g) 980 CaCO3 (s) → CaO (s) + CO₂ (g) 900 Kp value 4.1 x 10-4 50.0 1.67 1.04

Calculate AG°for each of the following reactions from the equilibrium constant at the temperature given. Reaction Temperature (°C) N₂ (g) + O₂ (g) → 2NO (g) 2000 H₂(g) + 1₂ (g) →→→ 2HI (g) 400 CO₂ (g) + H₂ (g) → CO (g) + H₂O (g) 980 CaCO3 (s) → CaO (s) + CO₂ (g) 900 Kp value 4.1 x 10-4 50.0 1.67 1.04

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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