What is the free energy change in kJ/mol for the process below at 43.9 °C when the concentration of A =0.88 M, B = 0.49 M and C = 0.69 M? g

Respuesta :

Answer:

-15.5 kJ/mol

Explanation:

2A ⇄ 2B + C

[tex]$ K = \frac{[C][B]^2}{[A]^2} $[/tex]

   = [tex]$ \frac{(0.69)(0.49)^2}{0.88^2} $[/tex]

  = 0.21

T = [tex]$ 43.9^{\circ} $[/tex]C

   = (273 + 43.9) K  =  316.9 K

[tex]$ \Delta G^{\circ} = -19.4 $[/tex] kJ/mol

R = 8.314 J/k-mol  =  0.008314 kJ/k-mol

[tex]$ \Delta G = \Delta G^{\circ} - RT \ln K $[/tex]

     [tex]$ =-19.4 - 0.008314 \times 316.9 \ln (0.21) $[/tex]

     = -15.5 kJ/ mol