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An aqueous mixture of phenol and ammonia has initial concentrations of 0.200 M C6H5OH(aq) and 0.120 M NH3(aq). At equilibrium, the C6H5O–(aq) concentration is 0.050 M. Calculate the equilibrium constant, K, for the reaction below.

C6H5OH(aq) + NH3(aq) C6H5O– + NH4+(aq)

Respuesta :

Answer:

The equilibrium constant, K, for the reaction = 0.23

Explanation:

Given that

               initial concentrations of phenol(C₆H₅OH) = 0.2 Molar

                                  & concentration of ammonia = 0.12 Molar

At equilibrium the phenoxide ion concentration (C₆H₅O⁻) = 0.05 Molar

                        C₆H₅OH(aq) + NH₃(aq)    ⇄     C₆H₅O⁻(aq) + NH₄⁺(aq)

Initial conc.         0.2               0.12                        0                    0

Equilibrium  (0.2 - 0.05)      (0.12 - 0.05)           0.05              0.05

                     = 0.15               = 0.07  

                According to law of mass action

                           [tex]K_{c} =\frac{[C_{6}H_{5}O^{-} ][NH_{4}^{+} ]}{[C_{6}H_{5}OH ][NH_{3} ]}[/tex]

                                 [tex]=\frac{0.05X0.05}{0.15X0.07}[/tex]

                                 [tex]= 0.23[/tex]

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We have that for the Question "Calculate the equilibrium constant, K, for the reaction C6H5OH(aq) + NH3(aq) C6H5O– + NH4+(aq)"

Answer:

  • Equilibrium constant, K = [tex]0.2381[/tex]

From the question we are told

An aqueous mixture of phenol and ammonia has initial concentrations of 0.2M [tex]C_6H_5OH_(aq)[/tex] and 0.120M[tex]NH_3(aq)[/tex]. At equilibrium, the [tex]C_6H_5O(aq)[/tex] concentration is 0.05M

At Equilibrium

[tex]C_6H_5OH- =0.2-0.050 =0.15M\\\\NH_3 =0.12-0.05=0.07\\\\C_6H_5O- = [NH_4+] =0.050[/tex]

Therefore,

[tex]K= \frac{[C_6H_5O-] [NH_4+]}{[C_6H_5OH][NH_3]}\\\\= \frac{0.050*0.05}{0.15*0.07}\\\\=0.2381[/tex]

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