Question

Calculate activation energy, a constant A of the Arrenius equation , Vant-Goff's coefficient and reaction constant at T3. Use formula of Vant-Hoff's law.

CO+H2O=CO2+H2,
k2, min^-1 mole^-1= 8.15*10^-3,
T2,K= 310,
k1, min^-1 mole^-1= 3.1*10^-4,
T1,K= 290,
T3,K= 300

Answer 1

Question #49802, Chemistry, Other

Calculate activation energy, a constant A of the Arrenius equation, Vant-Goff's coefficient and reaction constant at T3. Use formula of Vant-Hoff's law.

$$\begin{array}{l} \mathrm{CO} + \mathrm{H_2O} = \mathrm{CO_2} + \mathrm{H_2}, \\ \mathrm{k_2, min^{\wedge} - 1 \, mole^{\wedge} - 1 = 8.15 \cdot 10^{\wedge} - 3}, \\ \mathrm{T_2, K = 310}, \\ \mathrm{k_1, min^{\wedge} - 1 \, mole^{\wedge} - 1 = 3.1 \cdot 10^{\wedge} - 4}, \\ \mathrm{T_1, K = 290}, \\ \mathrm{T_3, K = 300} \end{array}$$

**Answer:**

the Arrenius equation:

$$k = A e^{-E/RT}$$

Vant-Hoff's law:

$$\gamma = (V_2 / V_1)^{10/(T_2 - T_1)}$$

in single mole concentration $V = k$

$$\ln \frac{k_1}{k_2} = \frac{E_a}{R} \cdot \left(\frac{1}{T_2} - \frac{1}{T_1}\right); \quad E_a = \frac{R \left(\ln k_1 - \ln k_2\right)}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}$$ $$\gamma = \left(\frac{(8.15 \cdot 10^{-3})}{3.1 \cdot 10^{-4}}\right)^{10/(310 - 290)} = 5.127$$ $$k_3 = k_1 \cdot \gamma^{(T_3 - T_1)/10} = 3.1 \cdot 10^{-4} \cdot 5.127^{(300 - 290)/10} = 15.89 \cdot 10^{-4}$$

**Energy of activation:**

$$E_a = 8.31 \cdot \ln \left(\frac{(8.15 \cdot 10^{\wedge} - 3)}{(3.1 \cdot 10^{\wedge} - 4)}\right) / \left(\frac{1}{310} - \frac{1}{290}\right) = -122115.933 \, J$$ $$A = e^{-E/RT^3} / k_3$$ $$A = \exp \left(122115.933 / (8.31 \cdot 300)\right) / (15.89 \cdot 10^{\wedge} - 4) = 1.18 \cdot 10^{24}$$

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