13>moles of N2=1-x
moles of H2=3-3x
moles of ammonia=2x
total moles=4-2x
so 2x/(4-2x)=16/100
get x and use
Kp=(2x)2*20(1-x)(3-3x)3*P
that gives the answer
1)For which of the following reversible reactions Kp may be equal to 0.5 atm and how?
a)PCl5 ---------------> PCl3 + Cl2
b)N2 + 3H2 ---------> 2NH3
c)2NO2 ---------------------> N2O4
2)A 1 litre container contains 2 moles of PCl5 initially. If at equilibrium Kc is found to be 1 the degree of dissociation for PCl5 is ? I am getting the answer as 1 but in the book its given as 1/2
3)The ammonia in equilibrium with a 1:3 N2 - H2 mixture at 20 atm and 700K amounts to 16%. Calculate Kp and Kc .I am getting Kp as 0.016 which is wrong :(
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7 Answers
11how r u getting ans to Q-2 as 1???
it comes out to be 0.732
and for 3 its 16% of wat????
11@ Euclid, I made some mistake so I was getting it as 1 for question 2
for question 3 read the question properly The ammonia in equilibrium with a 1:3 N2 - H2 mixture at 20 atm and 700K amounts to 16%
111)For which of the following reversible reactions Kp may be equal to 0.5 atm and how?
a)PCl5 ---------------> PCl3 + Cl2
b)N2 + 3H2 ---------> 2NH3
c)2NO2 ---------------------> N2O4
Answer to this is as follows
we have to see in which reaction when we will find out Kp we will get the unit as atm, its only in the option (a), so it is the correct answer
1357for 2
PCl5 = PCl3+Cl2
2-x x x
K=x^2/(2-x)=1
or x^2+x-2=0
so x=1
and degree of dissociation is x/2=1/2