Home
Class 11
CHEMISTRY
1.6 mol of PCl(5(g)) is placed in 4 dm""...

1.6 mol of `PCl_(5(g))` is placed in 4 dm`""^3` closed vessel. When the temperature is raised to 500 K, it decomposes and at equilibrium 1.2 mole of `PCl_(5(g))` remains. What is the `K_c` value for the decomposition of `PCl_(5(g))" to "Cl_(2(g))"at 500 K?"`

A

0.013

B

0.05

C

0.033

D

0.067

Text Solution

Verified by Experts

The correct Answer is:
C
1.6 mol of `PCl_5` is placed in `4 dm^3` closed vessel.
`underset("1.6 mol")(PCl_5) Leftrightarrow underset(0)(PCl_(3_(g)) +underset(0)(Cl_(2(g)) "Initial "`
`"(1.6 -x) mol x mol x mol (At equilibrium)"`
Given that 1.6-x=1.2
x=0.4 mol
Therefore, `[PCl_5]=(1/2)/4=0.3. [PCl_3]=(0.4)/(4)=0.1`
and `[Cl_2]=(0.4)/(4)=0.1 therefore K_(c)=([PCl_3] [Cl_2])/([PCl_5])=(0.1 xx 0.1)/(0.3)=0.033`
Promotional Banner

Similar Questions

Explore conceptually related problems

16 mol of PCl_(5)(g) is placed in 4 dm^(-3) closed vessel. When the temperature is raised to 500 K, it decompses and at equilibrium, 1.2 mol of PCl_(5)(g) remains. What is K_(c) value for the decomposition of PCl_(5)(g) to PCl_(3)(g) and Cl_(2)(g) at 500K.

For the reaction, PCl_(5)(g)to PCl_(3)(g)+Cl_(2)(g)

For the reaction : PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g) :

Four moles of PCl_(5) are heated in a closed 4 dm^(3) container to reach equilibrium at 400 K. At equilibrium 50% of PCl_(5) is dissociated. What is the value of K_(c) for the dissociation of PCl_(5) into PCl_(3) "and" Cl_(2) at 400 K ?

Initially, 0.8 mole of PCl_(5) and 0.2 mol of PCl_(30 are mixed in one litre vessel. At equilibrium, 0.4 mol of PCl_(3) is present. The value of K_(c ) for the reaction PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) would be

2.0 mole of PCl_(5) were nttoducedd in a vessel of 5.0 L capacity of a particular temperature At equilibrium, PCl_(5) was found to be 35 % dissociated into PCl_(3)and Cl_(2) the value of K_(c) for the reaction PCl_(3)(g)+Cl_(2)(g)hArrPCl_(5)(g)