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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

Vapour densit...

Question

Vapour density of $N_{2} O_{4}$ which dissciates according to teh equilibrium $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ is $25.67$ at $100^{o} C$ and a pressure of $1 a t m$ . Calculate the degree dissociatrion for the reaction.

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Solution

$N_{2} O_{4} - - - - \to N O_{2}$

Initially $N_{2} O_{4}$ has concentration=1

Then,at equilibrium $N_{2} O_{4}$ h has conc:=2x

S=92 V.D at eq.=30 and M.W.=60

$\frac{N_{i n i t i a l}}{N_{f i n a l}} = \frac{M_{f i n a l}}{M_{i n i t i a l}}$

$\frac{1}{1 - x} + 2 x = \frac{60}{92}$

$\frac{1}{1 + x} = \frac{60}{92}$

$X = \frac{92}{60} - 1$

$X = \frac{32}{60}$

X=0.53
Therefore, percentage of dissociation = $0.53 X 100 = 53$

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Vapour density of N2O4 which dissciates according to teh equilibrium N2O4(g)⇌2NO2(g) is 25.67 at 100oC and a pressure of 1 atm. Calculate the degree dissociatrion for the reaction.

N2O4−−−−→NO2Initially N2O4 has concentration=1Then,at equilibrium N2O4 h has conc:=2xS=92 V.D at eq.=30 and M.W.=60NinitialNfinal=MfinalMinitial11−x+2x=609211+x=6092X=9260−1X=3260X=0.53Therefore, percentage of dissociation =0.53X100=53

Vapour density of $N_{2} O_{4}$ which dissciates according to teh equilibrium $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ is $25.67$ at $100^{o} C$ and a pressure of $1 a t m$ . Calculate the degree dissociatrion for the reaction.