Question

Lanthanum(III) phosphate crystallizes as a hemihydrate, ${\mathrm{LaPO}}_4\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$ . When it is heated, it loses water to give anhydrous lanthanum(III) phosphate.

${\mathrm{LaPO}}_4\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)\to 2{\mathrm{LaPO}}_4\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

This reaction is first order in the chemical amount of The rate constant varies with temperature as follows

Calculate the activation energy of this reaction.

Short answer

The activation energy of the reaction is $6.6\mathrm{x}{10}^4{\mathrm{Jmol}}^{-1}$

Step-by-step solution

(a) Arrehinus eqaution:

The Arrehinius equation explains the relationship between activation energy and rate constant k and temparature T of the reaction.

$\mathrm{lnK}=\mathrm{lnA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}......\left(1\right)$

Where K is the rate constant,E_a is the activation energy,R is the universal gas constant, and k is the rate constant.

$$\begin{gathered} {\mathrm{lnk}}_1=\mathrm{lnA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_1}......\left(1\right) \\ {\mathrm{lnk}}_2=\mathrm{lnA}-\frac{{\mathrm{E}}_{\mathrm{a}}}{{\mathrm{RT}}_2}.......\left(2\right) \end{gathered}$$

Substraction equation (1) from (2)

$$\begin{gathered} {\mathrm{lnk}}_2-{\mathrm{lnk}}_1={\mathrm{E}}_{\mathrm{a}}\left(\frac{1}{{\mathrm{RT}}_1}-\frac{1}{{\mathrm{RT}}_2}\right) \\ \ln \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right).............\left(3\right) \end{gathered}$$

Activation energy

Substituting the given values

$$\begin{gathered} {\mathrm{k}}_1=2.3\mathrm{x}{10}^{-4}{\mathrm{s}}^{-1} \\ {\mathrm{T}}_1={205}^0\mathrm{C} \\ {\mathrm{T}}_1=\left(205+273\right)\mathrm{K} \\ {\mathrm{T}}_1=478\mathrm{K} \\ {\mathrm{k}}_2=3.69\mathrm{x}{10}^{-4} \\ {\mathrm{T}}_2={219}^0\mathrm{C} \\ {\mathrm{T}}_2=\left(219+273\right) \\ {\mathrm{T}}_2=492 \\ \mathrm{R}=8.315{\mathrm{Jmol}}^{-1}\mathrm{K} \end{gathered}$$

$$\begin{gathered} \mathrm{Substituting}\mathrm{these}\mathrm{values}\mathrm{in}\mathrm{eqaution}-3 \\ \ln \frac{3.69\mathrm{x}{10}^{-4}}{2.3\mathrm{x}{10}^{-4}}=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.315{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}}\left(\frac{1}{478\mathrm{K}}-\frac{1}{492\mathrm{K}}\right) \\ 0.472717=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.315{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}}\left(0.0020920-0.0020324\right)\mathrm{K} \\ 0.472717={\mathrm{E}}_{\mathrm{a}}\mathrm{x}0.0000071593366{\mathrm{J}}^{-1}\mathrm{mol} \\ {\mathrm{E}}_{\mathrm{a}}=\frac{0.472717}{0.0000071593366}{\mathrm{Jmol}}^{-1} \\ {\mathrm{E}}_{\mathrm{a}}=66028.094163501{\mathrm{Jmol}}^{-1} \\ {\mathrm{E}}_{\mathrm{a}}=6.6\mathrm{x}{10}^4{\mathrm{Jmol}}^{-1} \end{gathered}$$

Therefore activation energy of the reaction is $6.6\mathrm{x}{10}^4{\mathrm{Jmol}}^{-1}$