## Discussion on the Calculation of Rate Constant and Activation Energy of Second-Order Reaction

Li Qibiao, Hao Yajuan,

Abstract

In this paper, we discussed several problems on calculation of reaction rate constant in physical chemistry reference books. When calculating the rate constant k of the second-order reaction, we should make the half-life formula consistent with the reaction stoichiometric equation. And when calculating the activation energy with Arrhenius formula for ideal gas-phase reaction, attention should be paid to identifying the difference between the reaction rate constant kc and kp, as well as the difference between corresponding activation energy Eac and Eap. It is beneficial for students to have a correct understanding in calculation.

Keywords： Reaction rate constant ; Half-life ; Gas-phase reaction ; Activation energy

Li Qibiao. Discussion on the Calculation of Rate Constant and Activation Energy of Second-Order Reaction. University Chemistry[J], 2020, 35(9): 205-208 doi:10.3866/PKU.DXHX201910055

## 2 反应速率常数计算问题示例

${k_{p, 970{\rm{K}}}} = \frac{1}{{{p_{{{\rm{A}}_{\rm{0}}}}}{t_{1/2}}}} = \frac{1}{{39.2 \times {{10}^3}{\rm{ Pa}} \times 1529{\rm{ s}}}} = 1.7 \times {10^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}$

${k_{p, 1030{\rm{K}}}} = \frac{1}{{{p_{{{\rm{A}}_{\rm{0}}}}}{t_{1/2}}}} = \frac{1}{{48.0 \times {{10}^3}{\rm{ Pa}} \times 212{\rm{ s}}}} = 9.8 \times {10^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}$

${k_{p, 1030{\rm{K}}}} = \frac{1}{{2{p_{{{\rm{A}}_{\rm{0}}}}}{t_{1/2}}}} = 4.9 \times {10^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}$

$\ln \frac{{{k_p}_{(1030{\rm{K}})}}}{{{k_p}_{(970{\rm{K}})}}} = \frac{{{E_{\rm{a}}}}}{R}(\frac{1}{{970{\rm{ K}}}} - \frac{1}{{1030{\rm{ K}}}})$

$\ln \frac{{9.8 \times {{10}^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}}{{1.7 \times {{10}^{ - 8}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}} = \frac{{{E_{\rm{a}}}}}{{8.314{\rm{ J}} \cdot {\rm{mo}}{{\rm{l}}^{ - {\rm{1}}}} \cdot {{\rm{K}}^{ - {\rm{1}}}}}}(\frac{1}{{970{\rm{ K}}}} - \frac{1}{{1030{\rm{ K}}}})$

$\ln \frac{{{k_c}_{(1030{\rm{K}})}}}{{{k_c}_{(970{\rm{K}})}}} = \ln \frac{{{k_p}_{(1030{\rm{K}})} \cdot R \times 1030{\rm{ K}}}}{{{k_p}_{(970{\rm{K}})} \cdot R \times 970{\rm{ K}}}} = \frac{{{E_{{\rm{a}}c}}}}{R}(\frac{1}{{970{\rm{ K}}}} - \frac{1}{{1030{\rm{ K}}}})$

$\ln \frac{{{k_{c, {T_2}}}}}{{{k_{c, {T_1}}}}} = \frac{{{E_{{\rm{a}}c}}}}{R}(\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}})$

$\ln \frac{{{k_{p, {T_2}}}}}{{{k_{p, {T_1}}}}} = \frac{{{E_{{\rm{a}}p}}}}{R}(\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}})$

${E_{{\rm{a}}c}} - {E_{{\rm{a}}p}} = (n - 1)R\frac{{{T_2}{T_1}}}{{{T_2} - {T_1}}} \cdot \ln \frac{{{T_2}}}{{{T_1}}}$

(1) ${K_p} = \frac{{{k_{p, 1}}}}{{{k_{p, - 2}}}} = \frac{{{p_{\rm{B}}}{p_{\rm{C}}}}}{{{p_{\rm{A}}}}} = \frac{{0.21{\rm{ }}{{\rm{s}}^{ - {\rm{1}}}}}}{{5 \times {{10}^{ - 9}}{\rm{ P}}{{\rm{a}}^{ - {\rm{1}}}} \cdot {{\rm{s}}^{ - {\rm{1}}}}}} = 4.2 \times {10^7}{\rm{ Pa}}$

(2)由于${E_{{\rm{a}}, 1}}{\rm{ = }}{E_{{\rm{a}}, - 2}}{\rm{ = }}{E_{\rm{a}}}$

$\ln \frac{{{k_{{T_2}}}}}{{{k_{{T_1}}}}}{\rm{ = }}\ln 2 = \frac{{{E_{\rm{a}}}}}{R}(\frac{1}{{298{\rm{ K}}}} - \frac{1}{{310{\rm{ K}}}})，{E_{\rm{a}}} = 44.36{\rm{ kJ}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}。$

(3)根据平衡常数与温度的关系式，在相同的升温区间内正、逆反应速率常数的变化值相同，所以

$\frac{{{\rm{d}}\ln {K_p}}}{{{\rm{d}}T}} = \frac{{{\Delta _{\rm{r}}}{H_{\rm{m}}}}}{{R{T^2}}} = 0，{\Delta _{\rm{r}}}{H_{\rm{m}}} = {\rm{0}}$

${\Delta _{\rm{r}}}{U_{\rm{m}}} = {\Delta _{\rm{r}}}{H_{\rm{m}}} - \sum {{\nu _{\rm{B}}}RT}$

T = 298 K时，ΔrUm = −2.48 kJ·mol−1T = 308 K时，ΔrUm = −2.56 kJ·mol−1

$\ln \frac{{{k_{{c_1}, 310{\rm{K}}}}}}{{{k_{{c_1}, 298{\rm{K}}}}}} = \ln \frac{{{k_{{p_1}, 310{\rm{K}}}}}}{{{k_{{p_1}, 298{\rm{K}}}}}} = \ln 2 = \frac{{{E_{{\rm{a}}c, 1}}}}{R}(\frac{1}{{298{\rm{ K}}}} - \frac{1}{{310{\rm{ K}}}})$

$\ln \frac{{{k_{{c_{ - 2}}, 310{\rm{K}}}}}}{{{k_{{c_{ - 2}}, 298{\rm{K}}}}}} = \ln \frac{{{k_{{p_{ - 2}}, 310{\rm{K}}}}R \times 310{\rm{ K}}}}{{{k_{{p_{ - 2}}, 298{\rm{K}}}}R \times 298{\rm{ K}}}} = \ln 2 + \ln \frac{{310{\rm{ K}}}}{{298{\rm{ K}}}} = \frac{{{E_{{\rm{a}}c, - 2}}}}{R}(\frac{1}{{298{\rm{ K}}}} - \frac{1}{{310{\rm{ K}}}})$

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