Ask your own question, for FREE!
Chemistry 6 Online
OpenStudy (anonymous):

in a 1 litre vessel at 1000k are introduced 0.1 mole each of NO and Br2 and 0.01 mole of NOBr 2NO(g)+Br2(g)=2NOBr(g) Kc=1.32*10^-2 determine the direction of the net reaction and calculate the partial pressure of NOBr in the vessel at equilibrium?

OpenStudy (anonymous):

@UnkleRhaukus @timo86m @ghazi

OpenStudy (anonymous):

hlp me out

OpenStudy (ghazi):

\[K _{p}=\frac{ P _{NoBr} }{ [P _{NO}]*[P _{Br2}] }\]

OpenStudy (ghazi):

and use Kp and Kc relation before this

OpenStudy (anonymous):

can u elaborate it

OpenStudy (anonymous):

bt by this i will nt gt the answer.

OpenStudy (anonymous):

hey plz hlp me out

OpenStudy (anonymous):

plz hlp me out

OpenStudy (anonymous):

@saifoo.khan

OpenStudy (ghazi):

\[Kp=Kc*(RT)^{\Delta n}\]

OpenStudy (ghazi):

now to find Kc you have been provided with everything

OpenStudy (anonymous):

Kc is given

OpenStudy (anonymous):

@DLS plz hlp me out

OpenStudy (anonymous):

@DLS can u hlp me out

OpenStudy (anonymous):

@lalaly

OpenStudy (anonymous):

hlp me out.

OpenStudy (unklerhaukus):

i can not do these

OpenStudy (anonymous):

i know

OpenStudy (anonymous):

Find out the reaction quotient

OpenStudy (anonymous):

can u ? do

OpenStudy (anonymous):

i dunno much more

OpenStudy (anonymous):

reaction quotient? i'm nt getting wati t mean

OpenStudy (anonymous):

(NOBr)^2 / (NO)^2 * (Br2)

OpenStudy (anonymous):

if its greater than kc, move backward, less than means forward

OpenStudy (anonymous):

oh i knew after this

OpenStudy (anonymous):

pressure i dunno how to do

OpenStudy (anonymous):

can u eloborate it

OpenStudy (anonymous):

i dont know the pressure part

OpenStudy (anonymous):

hey plz hlp me out

OpenStudy (anonymous):

@saifoo.khan

OpenStudy (anonymous):

oh sorry i wouldn't know

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

@TuringTest can u hlp me out

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!