Question

Phosphorus pentachloride decomposes to phosphorus trichloride at
high temperatures according to the equation:

PCl_{5}(*g*) ↔ PCl_{3}(*g*) +
Cl_{2}(*g*)

At 250° 0.188 M PCl_{5} is added to the flask. If
*K* _{c} = 1.80, what are the equilibrium
concentrations of each gas?

Phosphorus pentachloride decomposes to phosphorus trichloride at
high temperatures according to the equation:

PCl_{5}(*g*) ↔ PCl_{3}(*g*) +
Cl_{2}(*g*)

At 250° 0.188 M PCl_{5} is added to the flask. If
*K* _{c} = 1.80, what are the equilibrium
concentrations of each gas?

[PCl_{5}] = 0.0164 M, [PCl_{3}] = 0.172 M, and
[Cl_{2}] = 0.172 M |

[PCl_{5}] = 1.80 M, [PCl_{3}] = 1.80 M, and
[Cl_{2}] = 1.80 M |

[PCl_{5}] = 0.0940 M, [PCl_{3}] = 0.411 M, and
[Cl_{2}] = 0.411 M |

[PCl_{5}] = 4.13 M, [PCl_{3}] = 3.94 M, and
[Cl_{2}] = 3.94 M |

Answer #1

PCl_{5} -----------------------------.>
PCl_{3} +
Cl_{2}

initial concentration 0.188 M 0 0

change in concentraion -x + x +x

equilibrium concentration 0.188-x + x +x

equilibrium constant , k_{c =} [PCl_{3}]
[Cl_{2}]/ [
PCl_{5}]

substitute the equillibriu conentration in the above formula

Kc = X . X/ 0.188-x

1.8 = x^{2}/ 0.188-x

x^{2} + 1.8 x - 0.3384 = 0

on solving the above quadratic we have x = +0.1716 ( negative value of x is rejected)

so the equilibrium concentrations of all gases :

[ PCl_{5}] = 0.188 - 0.1716

= 0.0164M

[ PCl_{3}] = 0.1716 M

[ Cl_{2}] = 0.1716M

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