IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry - a problem in chemical equilibrium.....please solve it

snehavenus

the ammonia in equilibrium with nitrogen-hydrogen mixturein the ratio 1:3 at 673K and a pressure of 4*10⁶Nm^-2 is 20% by volume. calculate k_p-(equilibrium constant in terms

of partial pressure)

Aatish

is the answer 6.63*10^-5

snehavenus

hi aatish,

please tell me how you did it.........i am sorry the answer given here is

5.789*10^-14 N^-2m⁴

sneha.bagri

intially let the no. of moles of N2 & H2 be 1 & 3 respectively

N2 +3H2 --> 2NH3

initially 1 3 0

equi. 1-x 3-3x 2x

% by vol. of gas =% by mole

2x/(4-2x)=0.2

x=1/3

mole fraction of H2=(3-3x)/(4-2x)=0.6

mole fraction of N2=(1-x)/(4-2x)=0.2

k_{p= ((mole fraction of NH3*totol pressure)^2)/((mol frac of}

_{N2 *total pressure)*(mol frac of H2 *total press)^3)}

_{=5.789*10^(-14)}

(2)

Hey sneha how did u get that step 2x/4-2x, it should be 2x/4-4x isnt it I think the anser is also quite wrong

Aatish

Hey 12345 !! 2x/4-2x is the mole fraction for ammonia

ie Mole fraction = No. of moles / Total no. of moles

well done sneha

arun-rashi

N2 +3H2 gives 2NH3 tatal p is 4 *10 to power 6,1-x 3-3x gives 2x ,Kp is cal by ratio of partial pressure which is mole frection *tatal pressure.total moles is 4-2x at equilibrium given volume percentage which is taken as mole per. beca P and T constent2x /4-2x =20/100 get x from here and calculate Kp.

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