Question

If the % of dissociation of n2o4 into no2 is 50, the ratio for kp and peq for n2o4 into 2no2 becomes equal to?

Answer 1

ratio of equilibrium constant to pressure is 2 : 3 .

for chemical reaction,

N2O4 ⇔2NO2

at t = 0, 1 0

at eq (1 - α) 2α

so, total = 1 - α + 2α = 1 + α

now P_(N2O4) = x_(N2O4) P

= (1 - α)(1 + α)P

P_(NO2) = 2α/(1 + α) P

K_p = P_(NO2)²/P_(N2O4)

={ 2α/(1 + α) P }²/{ (1 - α)(1 + α)P}

= 4α²P/(1 + α)

hence, equilibrium constant K_p = 4α²P/(1 + α)

here P is total pressure of gases

then, ratio of K_p and P is = K_p/P = 4α²/(1 + α)

now putting value of α = 50% = 0.5

so, K_p/P = 4(0.5)²/(1 + 0.5)

= 4(0.25)/1.5

= 1/1.5 = 2/3

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