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

-15.5 kJ/mol

Step-by-step explanation:

2A ⇄ 2B + C

`$ K = ([C][B]^2)/([A]^2) $`

=
`$ ((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 * 316.9 \ln (0.21) $`

= -15.5 kJ/ mol