The thermal decomposition rate can be expressed by Eq. (1) as a function of two independent variables: the transformation rate α and temperature *T* [19]:

$$\begin{array}{}{\displaystyle (\phantom{\rule{thinmathspace}{0ex}}\frac{d\alpha}{dt})T=k(T)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}f(\alpha )}\end{array}$$(1)

where *T* (*K*) is the absolute temperature of sample, *α* (%) is the decomposition rate, and *k* is the rate constant for the decomposition reaction. The heating rate *β* is constant in GTA, and DTA, and is expressed by Eq. (2):

$$\begin{array}{}{\displaystyle \beta =dT/dt}\end{array}$$(2)

The rate constant *k* depends on temperature *T* according to the Arrhenius equation [20]:

$$\begin{array}{}{\displaystyle k(\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}T)=A\mathrm{e}\phantom{\rule{thinmathspace}{0ex}}\mathrm{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{p}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}(\frac{{E}_{\alpha}}{RT})}\end{array}$$(3)

where *A* is the pre-exponential factor (*s*^{–1}), *E*_{a} is the activation energy (J·mol^{–1}), and *R* is the gas constant (8.314 J·mol^{–1}·K^{–1}).

Combining Eqs. (1), (2) and (3) provides Eq. (4), which is the rate law for the decomposition reaction:

$$\begin{array}{}{\displaystyle \frac{d\alpha}{dT}=\frac{1}{\beta}A\mathrm{e}\phantom{\rule{thinmathspace}{0ex}}\mathrm{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{p}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}(\frac{{E}_{\alpha}}{RT})f(\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\alpha )}\end{array}$$(4)

We calculated the kinetic triplet (the activation energy *E*_{a}, the pre-exponential factor *A*, and the mechanism function *g*(*α*)) for both steps of NPL-10 decomposition process.

The activation energy *E*_{a} was obtained by Eq. (5) using KAS iterative method:

$$\begin{array}{}{\displaystyle \text{1n}\frac{\beta}{h(x){T}^{2}}=\text{1n}[\frac{A{E}_{a}}{g(\alpha )R}]-\frac{{E}_{a}}{RT}}\end{array}$$(5)

where

$$\begin{array}{}{\displaystyle h(x)=\frac{{x}^{4}+18{x}^{3}+88{x}^{2}+96x}{{x}^{4}+20{x}^{3}+120{x}^{2}+240x+120}}\end{array}$$(6)

and

$$\begin{array}{}{\displaystyle x=\frac{{E}_{a}}{RT}}\end{array}$$(7)

KAS iterative method is divided into three steps: ① Assume *h*(*x*)=1 to estimate the initial activation energy *E*_{1} for given *β* and *g*(*α*) values by Eq. (5). ② Bring *E*_{1} and the corresponding *T* into Eq. (7), then use the calculated *x* value to derive *h*(*x*) by Eq. (6); enter *x* and *h*(*x*) values into Eq. (5), and use the linear regression of ln[*β*/*h*(*x*)*T*^{2}] *vs* 1/*T* to calculate the line slope (–*E*_{2}/*R*) and hence *E*_{2}. ③ Repeat step ② by replacing *E*_{1} by *E*_{2}. When |*E*_{i}–*E*_{M}| < 0.01 kJ·mol^{–1}, *E*_{i} can be considered a real value.

The activation energy *E*_{a} was calculated for *α* = 0.2–0.8 at 0.05-unit intervals (). A plot of the activation energy *E*_{a} as a function of *a* for steps 1 and 2 (Figure 4) shows two curves of different shape, indicating that decomposition follows a different mechanism in steps 1 and 2. The non-linearity of curves suggests a complex mechanism for both steps.

Figure 4 Line chart for thermal decomposition of NPL-10.

Table 2 Activation energies calculated by KAS iterative method.

To determine the most probable kinetic functior *g*(*α*) for the thermal decomposition of NPL-10, we performed the linear regressen anaiyais of kinetic functions () [18]. Eq. (5) was modified into Eq. (6):

Table 3 Linear regression analysis of 41 kinetic functions for steps 1 and 2.

$$\begin{array}{}{\displaystyle \text{1n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g(\alpha )=[\text{1n}\frac{AE}{R}+\text{1n}\frac{{e}^{-x}}{x}+\text{1n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p(x)]-\text{1n}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\beta \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(6)}\end{array}$$

where

$$\begin{array}{}{\displaystyle p(x)=\frac{{e}^{-x}}{{x}^{2}}h(x).}\end{array}$$

Linear regression analysis did not provide a straight line for four functions, namely 21 (dependent on pressure in nddhion to temperature, 25 (tew of Mampel Pewer), 39 and 40 (both functions of index law). The kinetic functions with the slope closest to –1 and best *R*^{2} values were *g*(*α*)= 1–(1–*α*)^{2} (, entry 33) for step 1 and *g*(*α*)= *α*^{1/2} for step 2 of thermal decomposition (, entry 31). These results indicate that the two steps of thermal decomposition follow a different order of reaction (*n*=2 for step 1 and *n*=1/2 for step 2).

Pre-exponential factors *A* were calculated by entering *E*_{a} values () and *g*(*α*) values () into Eq. (5). The pre-exponential factor of step 1 is *A*_{1}=3.86·10^{11}s^{–1}, the pre-exponential factor of step 2 is *A*_{2}=5.60·10^{13} s^{–1}. Calculation of *E*_{a}, *g*(*α*), and *A* allowed to derive the rate law of the decomposition process. The rate law for the first step of thermal decomposition is:

$$\begin{array}{}{\displaystyle \frac{d\alpha}{dT}=\frac{3.86\cdot {10}^{11}}{\beta}\text{exp}(-\phantom{\rule{negativethinmathspace}{0ex}}\frac{62.21\cdot {10}^{3}}{RT})\cdot [1-(1-\alpha {)}^{2}];}\end{array}$$

the rate law for the second step is:

$$\begin{array}{}{\displaystyle \frac{d\alpha}{dT}=\frac{5.60\cdot {10}^{13}}{\beta}\text{exp}(-\phantom{\rule{negativethinmathspace}{0ex}}\frac{75.16\cdot {10}^{3}}{RT})\cdot {\alpha}^{\frac{1}{2}}.}\end{array}$$

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