##### The equilibrium constant, K, for the following reaction is 5.37E-2 at 540 K. PCl

*label*Chemistry

*account_circle*Unassigned

*schedule*1 Day

*account_balance_wallet*$5

The equilibrium constant, K, for the following reaction is **5.37E-2** at **540** K.

**PCl**(g)

_{5}**PCl**(g) +

_{3}**Cl**(g)

_{2}An equilibrium mixture of the three gases in a

**18.5**L container at

**540**K contains

**0.167**M

**PCl**,

_{5}**9.48E-2**M

**PCl**and

_{3}**9.48E-2**M

**Cl**. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of

_{2}**9.64**L?

[PCl]_{5} | = | |

[PCl]_{3} | = | |

[Cl]_{2} | = |

It says you have to use the quadratic formula

Thank you for the opportunity to help you with your question!

PCl5= 5.203

PCl3=4.27

Cl2=0.16

wrong. look

**Step 1.** Determine the concentrations immediately after the volume has been reduced, before the system responds to the disturbance.

Rearranging the relationship M gives _{1}V_{1} = M_{2}V_{2}M_{2} = M_{1} | V_{1} |

V_{2} |

[PCl> = _{5}0.167 M | 18.5 L | = 0.320 M |

9.64 L |

[PCl> = _{3}9.48E-2 M | 18.5 L | = 0.182 M |

9.64 L |

[Cl> = _{2}9.48E-2 M | 18.5 L | = 0.182 M |

9.64 L |

**Step 2.** Predict the direction in which the reaction will proceed to reach equilibrium.

**PCl**(g) [img src="http://cxp.cengage.com/contentservice/assets/T=1432835347244/owms01h/mediaarchives/GenChem/Image/Doublearrow.GIF">

_{5}**PCl**(g) +

_{3}**Cl**(g) K =

_{2}**5.37E-2**

Le Chatelier's Principle predicts that reducing the volume of the system will cause the system to shift to the left, forming fewer moles of gas.

It is also possible to compare Q to K.

K = | [PCl>[_{3}Cl]_{2} | = 5.37E-2 |

[PCl]_{5} |

Q = | (0.182)(0.182) | = 0.103 > K |

(0.320) |

This also predicts a shift towards reactants as the system comes to equilibrium.

**Step 3.** Set up an ICE table to define equilibrium concentrations in terms of x.

**PCl**(g)

_{3}PCl (g)_{5} | + | Cl (g)_{2} | |

Initial (M) | 0.320 | 0.182 | 0.182 |

Change (M) | + x | - x | - x |

Equilibrium (M) | (0.320 + x) | (0.182 - x) | (0.182- x) |

**Step 4.** Substitute the equilibrium concentrations into the equilibrium constant expression and solve for **x**.

K = | [PCl>[_{3}Cl]_{2} | = | (0.182 - x)(0.182 -x ) | = 5.37E-2 |

[PCl]_{5} | (0.320 + x) |

Rearrange to get an expression of the form ax

^{2}+ bx + c = 0 and use the quadratic formula (see the information page) to solve for

**x**. This gives:

**x**=

**4.23E-2**,

**0.375**

The second value leads to results that are not physically reasonable.

**Step 5.** Use **x** = **4.23E-2** to solve for the final equilibrium concentrations.

[

**PCl**] =

_{5}**0.320**+

**4.23E-2**=

**0.363**M

[

**PCl**] =

_{3}**0.182**-

**4.23E-2**=

**0.140**M

[

**Cl**] =

_{2}**0.182**-

**4.23E-2**=

**0.140**M

**Step 6.** Check to see that the system is at equilibrium.

Q = | [PCl][_{3}Cl]_{2} | = | (0.140)(0.140) | = 5.37E-2 = K |

[PCl]_{5} | (0.363) |

*check_circle*

*check_circle*

*check_circle*

Secure Information

Content will be erased after question is completed.

*check_circle*