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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

Kc for, PCl...

Question

$K_{c}$ for, $P C l_{5} ⇌ P C l_{3} + C l_{2}$ is 13.7 mol litre $^{- 1}$ at $546 K$ . If the pressure developed in 10 litre box in equilibrium if reactant concentration is 1.0 mole is $P$ . The value of $100 P$ is_________.

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Solution

$P C l_{5} ⇋ P C l_{3} + C l_{2}$

Initial conc.(M) $0.1$ $0$ $0$
At eq. $0.1 - x$ $x$ $x$
The expression for the equilibrium constant is:

$K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]} = \frac{x \times x}{0.1 - x} = 13.7$
Hence, $x = 0.0993$

Total number of moles $= n = 10 (0.1 + x) = 10 (0.1 + 0.0993) = 1.993$

$P = \frac{n R T}{V} = \frac{1.993 \times 0.0821 \times 546}{10}$

$P = 8.93$

$100 P = 893$

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Similar questions

Q. Given that

K_{c} = 13.7

546 K

for

P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)

, calculate what pressure will develop in a

10

litre box at equilibrium at

546 K

when

1.00

mole of

P {C l}_{5}

is injected into the empty box?

Q. In the reaction

P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g)

the equilibrium concentration of

P C l_{5}

and

P C l_{3}

are 0.4 and 0.2 mole/litre respectively. If the value of

K_{C}

is 0.5 what is concentration of

C l_{2}

in moles/litre ?

Q. How much

P C l_{5}

must be added to a one-litre vessel at

250^{o} C

in order to obtain a concentration of 0.1 moles of

C l_{2}

K_{c}

for

P C l_{5} ⟺ P C l_{3} + C l_{2}

is 0.0414 mol/litre.

Q. X moles of

P C l_{5}

must be added to one-litre vessel at

250

K to obtain a concentration of 0.1 moles of

C l_{2}

K_{c}

for,

P C l_{5} ⇌ P C l_{3} + C l_{2}

, is

0.0414

mol / litre.

The value of 1000X is ____.

The equilibrium composition in 1 litre container for the reaction is
$P C l_{3} + C l_{2} ⇌ P C l_{5} 0.20 0.08 0.40$

0.22 mole of $C l_{2}$ is added at the same temperature and a new equilibrium is established. If the sum of the new equilibrium moles of $P C l_{3}, C l_{2} a n d P C l_{5}$ is p, then value of p is

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Kc for, PCl5⇌PCl3+Cl2 is 13.7 mol litre−1 at 546K. If the pressure developed in 10 litre box in equilibrium if reactant concentration is 1.0 mole is P. The value of 100P is_________.

PCl5⇋PCl3+Cl2Initial conc.(M) 0.1 0 0At eq. 0.1−x x xThe expression for the equilibrium constant is:Kc=[PCl3][Cl2][PCl5]=x×x0.1−x=13.7Hence, x=0.0993Total number of moles =n=10(0.1+x)=10(0.1+0.0993)=1.993P=nRTV=1.993×0.0821×54610P=8.93100P=893

$K_{c}$ for, $P C l_{5} ⇌ P C l_{3} + C l_{2}$ is 13.7 mol litre $^{- 1}$ at $546 K$ . If the pressure developed in 10 litre box in equilibrium if reactant concentration is 1.0 mole is $P$ . The value of $100 P$ is_________.