Home
Class 11
CHEMISTRY
AT673 K. in the formation of NH(3) From ...

AT673 K. in the formation of `NH_(3)` From `N_(2) and H_(2)` , the partial pressures of `N_(2) , H_(2) and and NH_(3)` at equilibrium are 0.5 , 1 and `9 xx 10^(-3) ` atm respectively. `(N_(2) + 3H_(2) hArr + Q)`
Calculate `K_(c)` for the reaction.

A

0.1

B

0.2

C

0.4

D

0.5

Text Solution

Verified by Experts

The correct Answer is:
D
`N_(2)+3H_(2) harr 2NH_(3)`
`K_(P)=(PN^(2)H_(3))/(PN_(2).PH_(2)^(3))=((9 xx 10^(-3))^(2))/(0.5 xx 1)=2 xx 81 xx 10^(-6)`
`K_(P)=K_(c)(RT)^(Delta n), K_(c)=(2 xx 81 xx 10^(-6))/((0.0821 xx 673)^(-2))=0.5`
Promotional Banner

Similar Questions

Explore conceptually related problems

AT673 K. in the formation of NH_(3) From N_(2) and H_(2) , the partial pressures of N_(2) , H_(2) and and NH_(3) at equilibrium are 0.5 , 1 and 9 xx 10^(-3) atm respectively. (N_(2) + 3H_(2) hArr + Q) Report Delta K_(C) Value

AT673 K. in the formation of NH_(3) From N_(2) and H_(2) , the partial pressures of N_(2) , H_(2) and and NH_(3) at equilibrium are 0.5 , 1 and 9 xx 10^(-3) atm respectively. (N_(2) + 3H_(2) hArr + Q) Report Delta G^(@) using K_(p) value

For the formation of NH_(3) from N_(2) and H_(2) at 500 K, the concentration of N_(2), H_(2) and NH_(3) at equilibrium are 1. 5xx10^(-2) M and 1.2xx10^(-2)M, respectively. The equilibrium constant for the reverse reaction is

For the formation of NH_3 from H_2 at 500 K, The concentration of N_2, H_2 and NH_3 at equilibrium are 1.5xx10^(-2) M , 3.0 x 10^(-2) M and 1.2 xx 10^(-2) M , respectively. The equilibrium constant for the reverse reaction is

For the reaction, N_(2) + 3H_(2) hArr 2NH_(3) in a vessel, equal moles of N_(2) and H_(2) are mixed to attain equilibrium.

For N_(2)+3H_(2)to2NH_(3) rates of disappearance of N_(2) and H_(2) and rate of appearance of NH_(3) respectively, are a,b and c then