### TranscriptionThe equilibrium constant for the reaction 2 HF (g) ⇌ H₂ (g) + F₂ (g) is 0.150 at a particular temperature. What is the equilibrium constant for the equation ½ H₂ (g) + ½ F₂ (g) ⇌ HF (g)?### ExplanationThe image displays a scientific question about chemical equilibrium constants. A virtual calculator is adjacent to the question, allowing for calculation. The equilibrium constant given is for the reaction involving hydrogen fluoride (HF) forming hydrogen (H₂) and fluorine (F₂) gases. The task is to find the equilibrium constant for the reverse and modified reaction where half a mole of H₂ and half a mole of F₂ form HF gas.

Chemical Reaction :- 2 HF (g) ⇌ H2 (g) + F2 (g) Kc = 0.150 1/2 H2 (g) + 1/2 F2 (g) ⇌ HF (g)…

Answered: The equilibrium constant for the…

The equilibrium constant for the reaction 2 HF (g) = H2 (g) + F2 (g) is 0.150 at a particular temperature. What is the equilibrium constant for the equation 2 H2 (g) + ½ F2 (g) = HF (g)?

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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### Transcription

The equilibrium constant for the reaction 2 HF (g) ⇌ H₂ (g) + F₂ (g) is 0.150 at a particular temperature. What is the equilibrium constant for the equation ½ H₂ (g) + ½ F₂ (g) ⇌ HF (g)?

### Explanation

The image displays a scientific question about chemical equilibrium constants. A virtual calculator is adjacent to the question, allowing for calculation. The equilibrium constant given is for the reaction involving hydrogen fluoride (HF) forming hydrogen (H₂) and fluorine (F₂) gases. The task is to find the equilibrium constant for the reverse and modified reaction where half a mole of H₂ and half a mole of F₂ form HF gas.

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