A student ran the following reaction in the laboratory at 282 K:\[ \text{2CH}_2\text{Cl}_2(\text{g}) \rightleftharpoons \text{CH}_4(\text{g}) + \text{CCl}_4(\text{g}) \]When she introduced \(6.98 \times 10^{-2}\) moles of \(\text{CH}_2\text{Cl}_2(\text{g})\) into a 1.00 liter container, she found the equilibrium concentration of \(\text{CH}_2\text{Cl}_2(\text{g})\) to be \(4.62 \times 10^{-3}\) M.Calculate the equilibrium constant, \(K_c\), she obtained for this reaction.\[ K_c = \underline{\phantom{aaaaaaaa}} \]

Given data,Initial moles of CH2Cl2=6.98×10-2molVolume of container=1.0LEquilibrium concentration of CH2Cl2=4.62×10-3M Initial concentration of CH2Cl2=Initial molesvolume=6.98×10-2mol1.0L=6.98×10-2M 2CH2Cl2(g) → CH4(g)+CCl4(g)Initial 6.98×10-2 - - Change -2x x xEquilibrium 6.98×10-2-2x x x6.98×10-2-2x=4.62×10-32x=6.98×10-2-4.62×10-32x=6.518×10-2x=3.26×10-2[CH4]=[CCl4]=xKc=[CH4][CCl4][CH2Cl2]2=3.26×10-2×3.26×10-2(4.62×10-3)2=10.6276×10-421.3444×10-6=0.498×102=49.8

Science

Chemistry

A student ran the following reaction in the laboratory at 282 K: 2CH,Cl½(g) CH4(g) + CC14(g) When she introduced 6.98×10-² moles of CH,Cl(g) into a 1.00 liter container, she found the equilibrium concentration of CH,Cl2(g) to be 4.62×103 M. Calculate the equilibrium constant, K., she obtained for this reaction. K. =

A student ran the following reaction in the laboratory at 282 K: 2CH,Cl½(g) CH4(g) + CC14(g) When she introduced 6.98×10-² moles of CH,Cl(g) into a 1.00 liter container, she found the equilibrium concentration of CH,Cl2(g) to be 4.62×103 M. Calculate the equilibrium constant, K., she obtained for this reaction. 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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