Question

kp for the reaction N2O4 to 2NO2 is 0.157 atm at 27°C and 1 atm pressure .calculate the value of kc
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Answer 1

Answer:

N2O4⇋2NO2

KP=0.66

Peq=0.5 atm

            N2O4⟺2NO2

At t=0        1            0

At t=eq     1−x    2x

Mole fraction of  N2O4=1+x1−x

Mole fraction of NO2=1+x2x

Partial pressure of N2O4=1+x1−x×0.5

Partial pressure of NO2=1+x2x×0.5

Kp=[P(N2O4)][P(NO)2]20.66=1

Answer 2

Given: Kp for the reaction $N_2O_4\rightarrow 2NO_2$ is 0.157 atm at 27°C.

To find: We have to find the value of Kc.

Solution:

We know that $Kp=Kc×(RT)^{∆n}$

Where Kp is the equilibrium constant and pressure=0.157 atm.

R is the Rydberg constant= 0.082lt.atm/(K.mol).

T is the temperature of the reaction.

T=27+273

T=300K.

∆n is the change in the number of moles.

For the reaction, $N_2O_4\rightarrow 2NO_2$ ∆n=1.

Putting the value in the above expression we get-

$Kp=Kc×(RT)^{∆n}\\0.157=Kc×(0.082×300)\\Kc=0.006$.

The value of Kc is 0.006.