Question

the moles of n2o4 and no2 at equilibrium are 1 and 2 respectively total pressure at equilibrium is 9 atm.find kp for the reaction n2o4(g)::::2no2(g)​

Answer 1

Answer:

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Answer 2

The value of $K_p$ for the reaction is, 12

Solution :  Given,

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Now we have to calculate the partial pressure of $N_2O_4\text{ and }NO_2$

$\text{ Partial pressure of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Total number of moles}}\times P=\frac{1}{3}\times 9=3atm$

$\text{ Partial pressure of }NO_2=\frac{\text{Moles of }NO_2}{\text{Total number of moles}}\times P=\frac{2}{3}\times 9=6atm$

Now we have to calculate the value of $K_p$ for the reaction.

The expression of $K_p$ will be,

$K_p=\frac{(p_{NO_2})^2}{p_{N_2O_4}}$

$K_p=\frac{(6)^2}{3}$

$K_p=12$

Thus, the value of $K_p$ for the reaction is, 12

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