Question

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

Answer 1

Answer:

-15.5 kJ/mol

Explanation:

2A ⇄ 2B + C

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

   = $\frac{(0.69)(0.49)^2}{0.88^2}$

  = 0.21

T = $43.9^{\circ}$C

   = (273 + 43.9) K  =  316.9 K

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

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

$\Delta G = \Delta G^{\circ} - RT \ln K$

     $=-19.4 - 0.008314 \times 316.9 \ln (0.21)$

     = -15.5 kJ/ mol