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Relative Lowering of Vapour Pressure

P Cl 5⇌ P Cl ...

Question

$P C l_{5} ⇌ P C l_{3} + C l_{2}$ in the reversible reaction, the moles of $P C l_{5}$ , $P C l_{3}$ and $C l_{2}$ are a, b and c respectively at equilibrium and total pressure is P, then value of $K_{p}$ is:

$\frac{b c}{a} R T$

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$\frac{b}{(a + b + c)} P$

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$\frac{b c P}{a (a + b + c)}$

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$\frac{c}{(a + b + c)} P$

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Solution

The correct option is C
$\frac{b c P}{a (a + b + c)}$

$P C l_{5} ⇌ P C l_{3} + C l_{2}$

$MolesabcMole fractiona(a+b+c)b(a+b+c)c(a+b+c)$

Partial pressure: $\frac{a}{(a + b + c)} . P \frac{b}{(a + b + c)} . P \frac{c}{(a + b + c)} . P$

$K_{p} = \frac{\frac{b}{(a + b + c)} . P \frac{c}{(a + b + c)} . P}{\frac{a}{(a + b + c)} . P}$

$= \frac{b c}{a (a + b + c)} P$

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Similar questions

$P C l_{5} ⇌ P C l_{3} + C l_{2}$ in the reversible reaction, the moles of $P C l_{5}$ , $P C l_{3}$ and $C l_{2}$ are a, b and c respectively at equilibrium and total pressure is P, then value of $K_{p}$ is:

P {C l}_{5} ⇌ P {C l}_{3} + {C l}_{2}

in the reversible reaction the moles of

P {C l}_{5}, P {C l}_{3}, {C l}_{2}

are

a, b

and

c

respectively and total pressure is

P

then value of

K_{p}

is:

P C l_{5} (g)

\to

P C l_{3} (g) + C l_{2} (g)

in the reversible reaction, the moles of

P C l_{5}, P C l_{3} a n d C l_{2}

are a, b and c respectively and total pressure at equilibrium is P, then value of

K_{p}

is:

P C l_{3} ⇌ P C l_{3} + C l_{2}

in the reversible reaction the moles of

P C l_{3}

and

C l_{2}

are

a, b

and

c

respectively and total pressure is

P

, then value of

K_{p}

is:

Q. A reaction at equilibrium involving 2 moles each of

P C l_{5}, P C l_{3},

and

C l_{2}

is maintained at

250^{o} C

and a total pressure of 3 atm. The value of

K_{p}

is :
Given:

P C l_{5} ⇌ P C l_{3} + C l_{2}

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PCl5⇌PCl3+Cl2 in the reversible reaction, the moles of PCl5, PCl3 and Cl2 are a, b and c respectively at equilibrium and total pressure is P, then value of Kp is:

The correct option is C bcPa(a+b+c) PCl5⇌PCl3+Cl2 MolesabcMole fractiona(a+b+c)b(a+b+c)c(a+b+c) Partial pressure: a(a+b+c).Pb(a+b+c).Pc(a+b+c).P Kp=b(a+b+c).Pc(a+b+c).Pa(a+b+c).P =bca(a+b+c)P

$P C l_{5} ⇌ P C l_{3} + C l_{2}$ in the reversible reaction, the moles of $P C l_{5}$ , $P C l_{3}$ and $C l_{2}$ are a, b and c respectively at equilibrium and total pressure is P, then value of $K_{p}$ is: