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# The equilibrium constant (Kp) is 0.14 at a particular temperature for the reaction: N204(9) : 2N02(9) Given the following sets of initial conditions,...

Can someone help me with my chemistry questions? I'm confused. Thanks

The equilibrium constant (Kp) is 0.14 at a particular temperature for the reaction:
N204(9) : 2N02(9) Given the following sets of initial conditions, what is the net change that must occur for the reaction to reach equilibrium? Does the reaction shift left to reach equilibrium, does the reaction shift right to reach equilibrium or is the reaction at
equilibrium at these initial concentrations so no net change will occur? PM); = 0.142 atm, PNzo‘ = 0.144 atm righl Pm)2 = 0.111 atm, PN204 = 0.146 atm left a PM;2 = 0.065 atm, PN204 = 0.066 atm
ﬁt Pm)2 = 0.200 atm, PNZQ4 = 0.13 atm reguilibrriurn Pm;2 = 0.089 atm, PN204 = 0.056 atm Submiunsvger &quot;Tries 8/15 militias

The equilibrium constant is 8.8 x 102 at a particular temperature for the reaction: “2(9) +1209): 2H1(9) Use this information to decide what will happen, given the following sets of initial conditions:
Will the reaction shift to the left, to the right or will the system be at equilibrium? PHz = 0.43 torr, Pyz = 0.058 torr, PH! = 5.93 torr A 2.0 L ﬂask contains 1.15 mol H2, 0.077 mol 1; and 8.27 mol HI. , PH2 = 0.5293 torr,P12 = 0.07 torr, PHI = 5.71 torr
c . as contains . mu 1, . mo 2 an 1 mo .
A40Lfl k ' 00458 IH 0011 I! d0666 IHI submit Answer Tries 0/ 15

For que - 1 Q = 0.14 -------- is at equilibrium. Q = 0.084 ------------ will shift right. Q = 0.064 -------------will shift... View the full answer

1 comment
• Here sentences above Keq/KP implies Keq or KP, do not get confused with divide sign.
• ankurjainiitk
• May 03, 2018 at 9:16pm

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