Consider the equilibrium system described by the chemical reaction below. For this reaction, Kp = 109 at 298 K. If the equilibrium partial pressures of Br. and NOBR at 0.0159 atm and 0.0768 atm, respectively, determine the partial pressure of NO at equilibrium.. 2 NO(g) + Br:(g) =2 NOBr(g) NEXT (x) represents the equilibrium partial pressure of NO, set up the equilibrium expression for Kp to plve for the partial pressure. Do not combine or simplify terms.. 109 %3D O RESET (109) (0.0159) (0.0768) 2(0.0159) 2(0.0768) (0.0159) (0.0768) (x) (2x) (x) (2x)

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Consider the equilibrium system described by the chemical reaction below. For this
reaction, Kp = 109 at 298 K. If the equilibrium partial pressures of Br. and NOBR at
0.0159 atm and 0.0768 atm, respectively, determine the partial pressure of NO at
equilibrium..
2 NO(g) + Br:(g) = 2 NOBr(g)
2
NEXT
If (x) represents the equilibrium partial pressure of NO, set up the equilibrium expression for Kp to
solve for the partial pressure. Do not combine or simplify terms..
Kp
109
%3D
RESET
(109)
(0.0159)
(0.0768)
20.0159)
2(0.0768)
(0.0159)*
(0.0768)
(x)
(2x)
(2x)
Transcribed Image Text:Consider the equilibrium system described by the chemical reaction below. For this reaction, Kp = 109 at 298 K. If the equilibrium partial pressures of Br. and NOBR at 0.0159 atm and 0.0768 atm, respectively, determine the partial pressure of NO at equilibrium.. 2 NO(g) + Br:(g) = 2 NOBr(g) 2 NEXT If (x) represents the equilibrium partial pressure of NO, set up the equilibrium expression for Kp to solve for the partial pressure. Do not combine or simplify terms.. Kp 109 %3D RESET (109) (0.0159) (0.0768) 20.0159) 2(0.0768) (0.0159)* (0.0768) (x) (2x) (2x)
Consider the equilibrium system described by the chemical reaction below. For this
reaction, Kp = 109 at 298 K. If the equilibrium partial pressures of Brz and NOBR at
0.0159 atm and 0.0768 atm, respectively, determine the partial pressure of NO at
equilibrium..
%3D
2 NO(g) + Br:(g) =2 NOBr(g)
PREV
Based on your expression for Kp, solve for the equilibrium partial pressure of NO.
PNO
atm
%3D
RESET
0.0443
6.36
526
1.70 x 10-
0.0583
Transcribed Image Text:Consider the equilibrium system described by the chemical reaction below. For this reaction, Kp = 109 at 298 K. If the equilibrium partial pressures of Brz and NOBR at 0.0159 atm and 0.0768 atm, respectively, determine the partial pressure of NO at equilibrium.. %3D 2 NO(g) + Br:(g) =2 NOBr(g) PREV Based on your expression for Kp, solve for the equilibrium partial pressure of NO. PNO atm %3D RESET 0.0443 6.36 526 1.70 x 10- 0.0583
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