At some temperature and under a pressure of 4 atm, $P C l_{5}$ is 10% dissociated. Calculate the pressure at which $P C l_{5}$ will be 20% dissociated, temperature remaining same.
19.2 atm
0.192 atm
0.096 atm
0.96 atm

Question

At some temperature and under a pressure of 4 atm, $P C l_{5}$ is 10% dissociated. Calculate the pressure at which $P C l_{5}$ will be 20% dissociated, temperature remaining same.
19.2 atm
0.192 atm
0.096 atm
0.96 atm

Toppr Admin · Accepted Answer

At 4 atm pressure and 10- dissociation-PCl5 -x2192-xA0- -xA0- -xA0- -xA0- PCl3-xA0- -xA0- -xA0- -xA0- Cl2Equibm mole-xA0-xA0- -xA0-90-xA0- -xA0- -xA0- -xA0- -xA0- 10 -xA0-xA0- -xA0- -xA0- -xA0-10Equilibrium partial pressures90110-xD7-4-xA0- -xA0- -xA0- 10110-xD7-4-xA0- -xA0- -xA0- -xA0-10110-xD7-4Hence- Kp-PPCl3PCl2PPCl5-499-

	$P C l_{5}$ $\to$	$P C l_{3}$	$C l_{2}$
Equibm mole	90	10	10
Equilibrium partial pressures	$\frac{90}{110} \times 4$	$\frac{10}{110} \times 4$	$\frac{10}{110} \times 4$

At some temperature and under a pressure of 4 atm, PCl5 is 10% dissociated. Calculate the pressure at which PCl5 will be 20% dissociated, temperature remaining same.19.2 atm0.192 atm0.096 atm0.96 atm

At 4 atm pressure and 10% dissociation.PCl5 → PCl3 Cl2Equibm mole 90 10 10Equilibrium partial pressures90110×4 10110×4 10110×4Hence, Kp=PPCl3PCl2PPCl5=499.

At some temperature and under a pressure of 4 atm, $P C l_{5}$ is 10% dissociated. Calculate the pressure at which $P C l_{5}$ will be 20% dissociated, temperature remaining same.
19.2 atm
0.192 atm
0.096 atm
0.96 atm

At 4 atm pressure and 10% dissociation.

$P C l_{5}$ $\to$
$P C l_{3}$
$C l_{2}$
Equibm mole
90
10
10
Equilibrium partial pressures
$\frac{90}{110} \times 4$
$\frac{10}{110} \times 4$ $\frac{10}{110} \times 4$

Hence, $K_{p} = \frac{P_{P C l_{3}} P_{C l_{2}}}{P_{P C l_{5}}} = \frac{4}{99}$ .