help Chloroform, an important solvent, is preparedby the chlorination of methane at 721 K.CH4(g) + C12(g) → CH3CI(g) + HCI(g)What is the equilibrium constant for the reactionat 721 K?СH4(9) -74.6-50.5186.3Cl2(9)oO 0 223.1CH3CI(g) -81.9-60.2234.6HCI(g) -92.307 -95.299186.9

Chlorination of methane is given by CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) Temperature, T = 721 K…

Chloroform, an important solvent, is prepared by the chlorination of methane at 721 K. CH4(g) + Cl2(g) → CH3CI(g) + HCI(g) What is the equilibrium constant for the reaction at 721 K? СH4(9) -74.6 -50.5 186.3 Cl2(g) 0 o 223.1 СH3С(9) -81.9 -60.2 234.6 HCI(g) -92.307 -95.299 186.9

Chloroform, an important solvent, is prepared by the chlorination of methane at 721 K. CH4(g) + Cl2(g) → CH3CI(g) + HCI(g) What is the equilibrium constant for the reaction at 721 K? СH4(9) -74.6 -50.5 186.3 Cl2(g) 0 o 223.1 СH3С(9) -81.9 -60.2 234.6 HCI(g) -92.307 -95.299 186.9

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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Question

help

Chloroform, an important solvent, is prepared
by the chlorination of methane at 721 K.
CH4(g) + C12(g) → CH3CI(g) + HCI(g)
What is the equilibrium constant for the reaction
at 721 K?
СH4(9) -74.6
-50.5
186.3
Cl2(9)
o
O 0 223.1
CH3CI(g) -81.9
-60.2
234.6
HCI(g) -92.307 -95.299
186.9

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