Question

if Kp for the reaction N2O4=2NO2 is 0.66 then what is the equillibrium pressure of N2O4(total pressure at equillibrium is 0.5 atm)​

Answer 1

Answer:

N2O4<−−>2NO2

pressure initial: P

At equilibrium: P−X+2X

0.66=4X2/P−X-1

0.5=P+X-2

Solving 1 and 2 we get

x=0.16

y=0.34

2x=0.32=P(NO2)

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Answer 2

0.17 atm is the equilibrium pressure of$N_2O_4$.

Explanation:

Equilibrium pressure of $N_2O_4=p_1$

Equilibrium pressure of $NO_2=p_2$

$N_2O_4\rightleftharpoons 2NO_2$

The values of $K_p=0.66$

Total pressure at equilibrium = 0.5 atm

$p_1+p_2=0.5 atm$

$p_1=0.5-p_2$

$K_p=\frac{(p_2)^2}{(p_1)}$

$0.66=\frac{(p_2)^2}{(0.5-p_2)}}$

$p_2=0.33 atm$

$p-1=0.5 atm -0.33 atm = 0.17 atm$

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