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Chemistry 14 Online
OpenStudy (anonymous):

For this equilibrium reaction, which of the following describes how the given stress affects the equilibrium point? N2 (g) + O2 (g) ↔ 2NO (g) Answer Increase in pressure results in shifting of equilibrium to the right. Removal of NO results in shifting of equilibrium to the left. Increase in pressure results in shifting of equilibrium to the left. Removal of N2 results in shifting of equilibrium to the left.

OpenStudy (abhisar):

What doo you think should be the correct answer ?

OpenStudy (anonymous):

would it be the removal of N2 results in shifting of equilibrium to the left?

OpenStudy (abhisar):

Is more than one option correct ?

OpenStudy (anonymous):

multiple choice but only one answer

OpenStudy (anonymous):

why dont you tell me what YOU think since im the one asking for help lol

OpenStudy (abhisar):

ok i am confused between c and D...let's see what @abb0t says

OpenStudy (anonymous):

will he respond if he has not fanned me?

OpenStudy (abhisar):

yes i think he will respond.....

OpenStudy (abhisar):

looks like he is offline !

OpenStudy (anonymous):

he went offline... but you are between c and d? so you know it isnt a or be, correct?

OpenStudy (abhisar):

yes

OpenStudy (anonymous):

im nearly certain its d

OpenStudy (abhisar):

hmmm..then go for it :)

OpenStudy (abhisar):

@aryandecoolest

OpenStudy (anonymous):

i am confused between c and d.. :O

OpenStudy (abhisar):

BINGO ! XD

OpenStudy (abhisar):

Now i am sure both are correct !

OpenStudy (anonymous):

its d, thanks for at least narrowing it down haha

OpenStudy (abhisar):

are u sure its only d ?

OpenStudy (anonymous):

positive

OpenStudy (anonymous):

hahaha!! please explain how it's only d :)

OpenStudy (abhisar):

yeah...me too eagerly waiting !

OpenStudy (anonymous):

increasing the pressure would not shift the equilibrium but the removal of the N2 would. best way i can explain it lol. i do no the pressure increse would not affect the equilibrium

OpenStudy (abhisar):

Check this http://www.chemguide.co.uk/physical/equilibria/lechatelier.html

OpenStudy (abhisar):

`Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer molecules.`

OpenStudy (anonymous):

well ill be damned, but they were talking about increasing the pressure on the right side. witch would shift to the right side, not the left

OpenStudy (anonymous):

further into the problem witch i did not post was saying increase the right side of pressure

OpenStudy (abhisar):

There is no way we can increase the pressure of a particular side. Pressure is always increased or decreased for the whole reaction

OpenStudy (abhisar):

anyways i am still confused ! any one one among A and C has to be correct

OpenStudy (anonymous):

if you decrease the pressure the the equilibrium moves to the left and if increased it moves to the right

OpenStudy (abhisar):

and i think C is more apt......and D is also correct so both C and D is correct. That's my final conclusion.

OpenStudy (anonymous):

no see is wrong only because increasing the pressure would cause the equilibrium to shift to the right, not left.

OpenStudy (anonymous):

C

OpenStudy (abhisar):

In that case A should be correct...again two answers !

OpenStudy (anonymous):

oh well, i put d on my online quiz and was correct lol. thanks for the help!

OpenStudy (abhisar):

\(\rlap{\color{blue}{\huge\bigstar}\huge \color{blue}{ \text{You're Most Welcome! }}\color{blue}\bigstar}{\; \color{aqua}{\huge\bigstar}\huge \color{aqua}{\text{You're Most Welcome! }}\color{aqua}\bigstar}\) \(~~~~~~~~~~~~~~~~~~~~~~~~~~~\color{green}{\huge\ddot\smile}\color{blue}{\huge\ddot\smile}\color{pink}{\huge\ddot\smile}\color{red}{\huge\ddot\smile}\color{yellow}{\huge\ddot\smile}\)

OpenStudy (anonymous):

woah haha, thanks

OpenStudy (anonymous):

@abhisar how you make that :D teach me too dude :P

OpenStudy (abhisar):

@aryandecoolest oh well that's \(LaTeX\)....go to this link for Basic tutorial http://openstudy.com/study#/updates/52e12f56e4b0942cc9de719e

OpenStudy (anonymous):

thanx!!! )

OpenStudy (anonymous):

lol again......!!!

OpenStudy (abhisar):

What do you want to do ?

OpenStudy (anonymous):

write HELLO....what going wrong actually??

OpenStudy (abhisar):

``` \(\huge \cal \color{Blue}{Hello}\) ``` \(\huge \cal \color{Blue}{Hello}\)

OpenStudy (anonymous):

AHAA!!! what you did... are brackets going wrng for me? :P

OpenStudy (anonymous):

\(\huge \cal \color{GREEN}{THANKS}\)

OpenStudy (abhisar):

\(\rlap{\color{blue}{\huge\bigstar}\huge \color{blue}{ \text{You're Most Welcome! }}\color{blue}\bigstar}{\; \color{aqua}{\huge\bigstar}\huge \color{aqua}{\text{You're Most Welcome! }}\color{aqua}\bigstar}\) \(~~~~~~~~~~~~~~~~~~~~~~~~~~~\color{green}{\huge\ddot\smile}\color{blue}{\huge\ddot\smile}\color{pink}{\huge\ddot\smile}\color{red}{\huge\ddot\smile}\color{yellow}{\huge\ddot\smile}\)

OpenStudy (anonymous):

\(\huge \cal \color{red}{pleasure}\color{blue}{\ is}\color{green}{\ all}\color{orange}{\ mine}\)

OpenStudy (anonymous):

are yall done now

OpenStudy (abb0t):

If you ever have troubles, you can write an equilbirum expression, plug in some numbers and see how it will effect it.

OpenStudy (anonymous):

What is the binding energy for a C-12 nucleus with a mass defect of 0.0993 amu? [(1 amu= 1.66 × 10-27 ,1 J = 1 kg times m squared over s squared ) and the speed of light = 3.00 × 108 m over s ] Answer 1.48 × 10-11 J 1.24 × 10-12 J 4.12 × 10-21 J 8.94 × 1015 J @Abhisar

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