For reactions in the gas phase, an equilibrium constant may be written in terms of molarity (K) or in terms of partialpressures (Kp).The value of Ke for the reaction shown below is equal to 3.2x104 at 298 K.2NOBI (g) = 2NO(g)+Br,(g)M See Periodic Table O See HintWhat is the value for Kp at 298 Kfor the reaction represented by the equation shown below? The valuefor the gas constant, R, is 0.08206 L• atm/mol • K.NOB:(g)= NO (g) +-Br, (g)(3)-1g - (3)ON

When an equilibrium reaction is multiplied with a number then that number appears in the powers of…

For reactions in the gas phase, an equilibrium constant may be written in terms of molarity (K) or in terms of partial pressures (Ka). The value of Ke for the reaction shown below is equal to 3.2x104 at 298 K. 2NOBI(g) 2NO(g) +Br, (g) i See Periodic Table O See Hint What is the value for K, at 298 K for the reaction represented by the equation shown below? The value for the gas constant, R, is 0.08206 L• atm/mol • K. NOB (g) = NO(g) +Br, (g)

For reactions in the gas phase, an equilibrium constant may be written in terms of molarity (K) or in terms of partial pressures (Ka). The value of Ke for the reaction shown below is equal to 3.2x104 at 298 K. 2NOBI(g) 2NO(g) +Br, (g) i See Periodic Table O See Hint What is the value for K, at 298 K for the reaction represented by the equation shown below? The value for the gas constant, R, is 0.08206 L• atm/mol • K. NOB (g) = NO(g) +Br, (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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