\[ \require{physics} \require{mhchem} \require{ams} \]

レポート課題

Standard Gibss free energy のまとめ。

Compounds	\(\Delta G\ \text{KJ}\ \text{mol}^{-1}\)
Ba(s)	0
BaO(s)	-520.40592
BaS(s)	-456.056
BaSO4(s)	-1362.3104
O2(g)	0
S2(g)	80.08176

Show Code

import numpy as np

delta_G1 = -520.40592  # Ba(s) -> BaO(s)
delta_G2 = -456.056-(80.08176/2)   # Ba(s) -> BaS(s)
delta_G3 = -1362.3104-(80.08176/2)  # Ba(s) -> BaSO4(s)
delta_G4 = -1362.3104+520.40592-(80.08176/2)  # BaO(s) -> BaSO4(s)
delta_G5 = -1362.3104+456.056  # BaS(s) -> BaSO4(s)
delta_G6 = -520.40592+(80.08176/2)+456.056  # BaS(s) -> BaO(s) + S2(g)

delta_G1, delta_G2, delta_G3, delta_G4, delta_G5, delta_G6

(-520.40592,
 -496.09688,
 -1402.35128,
 -881.94536,
 -906.2544,
 -24.30904000000004)

\[ \begin{aligned} & \ce{Ba(s) + 1/2 O2(g) -> BaO(s)} \quad \Delta G_1 = -520.406\ \text{KJ mol}^{-1} \\ & \ce{Ba(s) + 1/2 S2(g) -> BaS(s)} \quad \Delta G_2 = -496.097\ \text{KJ mol}^{-1} \\ & \ce{Ba(s) + 1/2 S2(g) + 2O2(g) -> BaSO4(s)} \quad \Delta G_3 = -1402.35\ \text{KJ mol}^{-1} \\ & \ce{BaO(s) + 1/2 S2(g) + \frac{3}{2}O2(g) -> BaSO4(s)} \quad \Delta G_4 = -881.945\ \text{KJ mol}^{-1} \\ & \ce{BaS(s) + 2O2 -> BaSO4(s)} \quad \Delta G_5 = -906.254\ \text{KJ mol}^{-1} \\ & \ce{BaS(s) + \frac{1}{2}O2 -> BaO(s) + \frac{1}{2}S2(g)} \quad \Delta G_6 = -24.3090\ \text{KJ mol}^{-1} \\ \end{aligned} \]

\[ \begin{aligned} \frac{G_1}{RT} &= \frac{1}{2}\log P_{\ce{O_2}} \\ \frac{G_2}{RT} &= \frac{1}{2}\log P_{\ce{S_2}} \\ \frac{G_3}{RT} &= \frac{1}{2}\log P_{\ce{S_2}} + 2\log P_{\ce{O_2}} \\ \frac{G_4}{RT} &= \frac{1}{2}\log P_{\ce{S_2}} + \frac{3}{2}\log P_{\ce{O_2}} \\ \frac{G_5}{RT} &= 2\log P_{\ce{O_2}} \\ \frac{G_6}{RT} &= \frac{1}{2}\log P_{\ce{O_2}} - \frac{1}{2}\log P_{\ce{S_2}} \\ \end{aligned} \]

Show Code

import numpy as np
import matplotlib.pyplot as plt

# 定数設定
R = 8.314  # 気体定数 (J/(mol*K))
T = 900    # 温度 (K)

# ギブズ自由エネルギー変化 (J/mol)
delta_G1 = -520.406 * 1e3  # Ba -> BaO
delta_G2 = -496.097 * 1e3  # Ba -> BaS
delta_G3 = -1402.35 * 1e3  # Ba -> BaSO4
delta_G4 = -881.945 * 1e3  # BaO -> BaSO4
delta_G5 = -906.254 * 1e3  # BaS -> BaSO4
delta_G6 = -24.3090 * 1e3  # BaS -> BaO + S2

# 直線の描画
# y = log_PO2, x = log_PS2

# Ba -> BaO
x = np.linspace(-150, 0, 500)
log_PO2 = 2 * delta_G1 / (R * T)
y = np.full_like(x, log_PO2)

plt.plot(x, y, label='Ba -> BaO')

# Ba -> BaS
y = np.linspace(-150, 0, 500)
log_PS2 = 2 * delta_G2 / (R * T)
x = np.full_like(y, log_PS2)

plt.plot(x, y, label='Ba -> BaS')

# Ba -> BaSO4
x = np.linspace(-150, 0, 500)
y = -x / 4 + delta_G3 / (R * T) / 2

plt.plot(x, y, label='Ba -> BaSO4')

# BaO -> BaSO4
x = np.linspace(-150, 0, 500)
y = - x / 3 + delta_G4 / (R * T) / 3 * 2

plt.plot(x, y, label='BaO -> BaSO4')

# BaS -> BaSO4
x = np.linspace(-150, 0, 500)
log_PO2 = delta_G5 / (R * T) / 2
y = np.full_like(x, log_PO2)

plt.plot(x, y, label='BaS -> BaSO4')

# BaS -> BaO + S2
x = np.linspace(-150, 0, 500)
y = x  + delta_G6 / (R * T) * 2

plt.plot(x, y, label='BaS -> BaO + S2')

# Draw
plt.xlabel('log_PS2')
plt.ylabel('log_PO2')
plt.title('Graph of log_PS2 and log_PO2')
plt.legend()
plt.grid()
plt.show()

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