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Chemistry 18 Online
OpenStudy (vera_ewing):

How does adding NH3 to the reaction below affect the equilibrium of the reaction? N2(g) + 3H2(g) 2NH3(g) A. Increasing the amount of NH3 will make Q smaller than K. A net reaction occurs in the direction to increase NH3, or the forward reaction. B. Increasing the amount of NH3 will make Q smaller than K. A net reaction occurs in the direction to reduce NH3, or the reverse reaction. C. Increasing the amount of NH3 will make Q larger than K. A net reaction occurs in the direction to increase NH3, or the forward reaction. D. Increasing the amount of NH3 will make Q larger th

OpenStudy (vera_ewing):

@sweetburger A or B?

OpenStudy (sweetburger):

Does D say that Increasing the amount of NH3 will make Q larger than K. A net reaction occurs in the direction to reduce NH3, or the reverse reaction. ?

OpenStudy (vera_ewing):

Increasing the amount of NH3 will make Q larger than K. A net reaction occurs in the direction to reduce NH3, or the reverse reaction.

OpenStudy (sweetburger):

Pretty sure it is D. As q represents the initial reaction quotient. If the reaction quotient has the amount of products increased which is the numerator it will become larger than the K value. To get Q to become K a reverse reaction must occur.

OpenStudy (vera_ewing):

Yeah you're right. Thank you.

OpenStudy (vera_ewing):

OpenStudy (vera_ewing):

How about this one? I think either B or C.

OpenStudy (sweetburger):

i think you just take the square root of the solubility

OpenStudy (vera_ewing):

So D?

OpenStudy (sweetburger):

No, I believe it would be B.

OpenStudy (sweetburger):

i said that [Ag+][I-]=8.5x10^-17 then i let Ag and I represent x xo [x][x]=8.5x1-^-17 so \[\sqrt{8.5x10^-17} =9.2x10^-9\]

OpenStudy (sweetburger):

so x= 9.2x10^-9

OpenStudy (vera_ewing):

Ohh I see. Thanks for explaining that!

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