નીચે આપેલ માહિતી પરથી $H_2 + I_2\rightarrow 2HI$ પ્રક્રિયા માટે સક્રિયકરણ ઊર્જા .....

$T$ (in, $K$) $- 769$ , $1/T$ (in, $K^{-1}$ ) $- 1.3\times 10^{-3},$

$\log_{10}K - 2.9\,T$ (in, $K$) $- 667$, $1/T$ (in, $K^{-1}) - 1.5\times 10^{-3}$, $\log_{10}\,K - 1.1$

Question

નીચે આપેલ માહિતી પરથી $H_2 + I_2\rightarrow 2HI$ પ્રક્રિયા માટે સક્રિયકરણ ઊર્જા .....

$T$ (in, $K$) $- 769$ , $1/T$ (in, $K^{-1}$ ) $- 1.3\times 10^{-3},$

$\log_{10}K - 2.9\,T$ (in, $K$) $- 667$, $1/T$ (in, $K^{-1}) - 1.5\times 10^{-3}$, $\log_{10}\,K - 1.1$

Diffcult

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Solution

a
$\,\log \frac{{{K_2}}}{{{K_1}}}\, = \frac{{{E_a}}}{{2.303R}}\,\left( {\frac{{{T_2}\, - {T_1}}}{{{T_1}{T_2}}}} \right)$

$\log \frac{{{K_2}}}{{{K_1}}}\, = \frac{{{E_a}}}{{2.303R}}\left( {\frac{1}{{{T_1}}}\, - \,\frac{1}{{{T_2}}}} \right)$ $R = 2$

$\log {K_2}\, - \log {K_1}\, = \frac{{{E_a}}}{{2.303 \times 2}}[1.5 \times \,{10^{ - 3}}\, - \,1.3\,\, \times \,{10^{ - 3}}]$

$2.9 - 1.1 = \frac{{{E_a}}}{{2.303 \times 2}}\,\, \times 0.2 \times {10^{ - 3}}$

$1.8 = \frac{{{E_a}}}{{2.303 \times 2}} \times 0.2 \times {10^{ - 3}}$

${E_a} = \frac{{1.8\,\, \times \,\,2.303}}{{{{10}^{ - 4}}}}\,\, = \,\,4\,\, \times \,\,{10^4}$

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