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3 years ago

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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

1 answer:

nata0808 [166]3 years ago

8 0

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

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