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A mixture of PCl3 and Cl2 is allowed to react at 548 K. PCl3(g) + Cl2(g) PCl5(g) The initial concentration of the reactants are [PCl3] = 0.0641 M and [Cl2] = 0.0769 M. After the system reaches equilibrium, it is found that the Cl2 concentration has decreased to 0.0502 M. Based on these data, determine the value of the equilibrium constant, K, for this reaction.

Answer :

jufsanabriasa

Answer:

K = 14.2

Explanation:

For the reaction:

PCl₃(g) + Cl₂(g) ⇄ PCl₅(g)

The equilibrium constant, k, is defined as:

[tex]K = \frac{[PCl_5]}{[PCl_3][Cl_2]}[/tex](1)

Where concentrations are molar concentrations in equilibrium

When in the beginning there are just reactants, the reaction will occur until it reaches the equilibrium, the concentrations are:

PCl₃: 0.0641M - X

Cl₂: 0.0769-X

PCl₅: X

As the equlibrium concentration of Cl₂ is 0.0502M, X is:

Cl₂: 0.0769-X = 0.0502M; X = 0.0769M - 0.0502M = 0.0267M. Replaing, molar concentrations in equilibrium are:

PCl₃: 0.0641M - 0.0267M = 0.0374M

Cl₂: 0.0769- 0.0267M = 0.0502M

PCl₅: 0.0267M

Replacing in (1):

[tex]K = \frac{[0.0267]}{[0.0374][0.0502]}[/tex]

K = 14.2

I hope it helps!

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