At a certain temperature- the reaction PCl5g- PCl3g-Cl2g has an equilibrium constant Kc-5-8 - 10-2- Calculate the equilibrium concentrations of PCl5- PCl3 and Cl2 if only PCl5 is present initially at a concentration of 0-160M-

The correct option is A [PCl_3]=[Cl_2]=0.071 M, [PCl_5]=0.089 M #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; ...

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Standard XII

Chemistry

Soaps

At a certain ...

Question

At a certain temperature, the reaction $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ has an equilibrium constant $K_{c} = 5.8 \times 10^{- 2}$ . Calculate the equilibrium concentrations of $P C l_{5}, P C l_{3}$ and $C l_{2}$ if only $P C l_{5}$ is present initially at a concentration of $0.160$ M.

[P C l_{3}] = [C l_{2}] = 0.071 M, [P C l_{5}] = 0.089 M

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[P C l_{3}] = 0.034 M, [C l_{2}] = 0.071 M, [P C l_{5}] = 0.056 M

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[P C l_{3}] = 0.075 M [C l_{2}] = 0.088 M, [P C l_{5}] = 0.099 M

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[P C l_{3}] = [C l_{2}] = 0.043 M, [P C l_{5}] = 0.089 M

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Solution

The correct option is A $[P C l_{3}] = [C l_{2}] = 0.071 M, [P C l_{5}] = 0.089 M$
$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$
Initial conc. $0.16$
At eqm. $0.16 - x$ $x$ $x$
$K_{c} = \frac{x \times x}{(0.16 - x)} = 0.058$
$x = [P C l_{3}] = [C l_{2}] = 0.071 M$
$[P C l_{5}] = 0.089 M$

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Similar questions

Q. For

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g),

write the expression of

K_{c} .

Q. An equilibrium mixture of

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

and

{C l}_{2}

at a certain temperature contians

0.5 \times 10^{- 3} M

P {C l}_{3}

1.5 \times 10^{- 2}

P {C l}_{3}

and

3 \times 10^{- 2} M

{C l}_{2}

. Calculate the equilibrium constant

K_{c}

for the reaction

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

Q. 2 mole each of

P C l_{5}

and

C l_{2}

and 1 mole of

P C l_{3}

are taken in 5lt. flask at

27^{o} C

. After following equilibrium is attained in gaseous phase,

P C l_{5} ⇌ P C l_{3} + C l_{2}, P C l_{5}

is found to be 40% dissociated. Calculate equilibrium constant

K_{c}

Q. In the reaction

P C l_{5} ⇌ P C l_{3} + C l_{2}

the partial pressure of

P C l_{3}, C l_{2}

and

P C l_{5}

are

0.3, 0.2

and

0.6 a t m

respectively at equilibrium. If partial pressure of

P C l_{3}

and

C l_{2}

was increased twice, what will be the partial pressure of

P C l_{5}

is in

a t m

at new equilibrium condition:

Q. An equilibrium mixture of

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

and

C l_{2}

at a certain temperature contains

0.5 \times 10^{- 3} M P C L_{5}

1.5 \times 10^{- 2} M P C L_{3}

and

3 \times 10^{- 2} M C l_{2}

.Calculate the equilibrium constant

K_{c}

for the reaction

P C L_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

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At a certain temperature, the reaction PCl5(g)⇌PCl3(g)+Cl2(g) has an equilibrium constant Kc=5.8×10−2. Calculate the equilibrium concentrations of PCl5,PCl3 and Cl2 if only PCl5 is present initially at a concentration of 0.160M.

The correct option is A [PCl3]=[Cl2]=0.071M,[PCl5]=0.089M PCl5(g)→PCl3(g)+Cl2(g)Initial conc. 0.16At eqm. 0.16−x x xKc=x×x(0.16−x)=0.058x=[PCl3]=[Cl2]=0.071M[PCl5]=0.089M

The correct option is A $[P C l_{3}] = [C l_{2}] = 0.071 M, [P C l_{5}] = 0.089 M$
$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$
Initial conc. $0.16$
At eqm. $0.16 - x$ $x$ $x$
$K_{c} = \frac{x \times x}{(0.16 - x)} = 0.058$
$x = [P C l_{3}] = [C l_{2}] = 0.071 M$
$[P C l_{5}] = 0.089 M$