For the reaction : 2A(g) =3B(g) + C(g) ; Kp = 2.7 × 10–9 atm2 at 298 K. The degree of dissociation of A(g) at 298K and 40 atm is
(1) 0.1
(2) 0.01
(3) 0.001
(4) 0.0001
Answers
Answer:
(1) 0.1
Explanation:
N2O4⇌2NO2
Assume one mole of N2O4 is present initially.
0.310 moles of N2O4will dissociate to form 2×0.310=0.620 moles of NO2
1−0.310=0.690 moles of N2O4will remain.
Total pressure is 1 atm.
The partial pressure of N2O4
=0.690×1atm=0.690atm
The partial pressure of NO2
=0.620×1atm=0.620atm
Kp = PN2O4
PNO 2
2Kp= (0.690atm)(0.620atm)
2Kp=0.557
Total pressure is 10 atm.
Assume one mole of N2O4 is present initially.
x moles of N2O4
will dissociate to form 2x moles of NO2
.
1−x moles of N2O4 will remain.
Total pressure is 10 atm.
The partial pressure of N2O4
=(1−x)×10atm=(10−10x)atm
The partial pressure of NO2
=2x×10atm=20xatm
Kp = PN2O4
PNO2
20.557= ((10−10x)atm)(20xatm)
25.57−5.57x=400x2
400x 2+5.57x−5.57=0
This is quadratic equation with solution
x= 2a−b± b 2−4ac
x= 2(400)−(5.57)± (5.57) 2 −4(400)(−5.57)
x= 800−(5.57)± (5.57) 2−4(400)(−5.57)
x=0.111 or x=−0.125
The value x=−0.125 is discarded as the number of moles cannot be negative.
x=0.111
The degree of dissociation = 1.0.111
=0.111