$[$આપેલ છે :${R}=8.31\, {~J} \,{~K}^{-1} \,{~mol}^{-1} ; \log 6.36 \times 10^{-3}=-2.19$ $\left.10^{-4.79}=1.62 \times 10^{-5}\right]$
\({~K}_{600}=x \times 10^{-6} {~s}^{-1}\)
\({E}_{{a}}=209 {~kJ} / {mol}\)
Applying ;
\(\log \left(\frac{{K}_{{T}_{2}}}{{~K}_{{T}_{1}}}\right)=\frac{-{E}_{{a}}}{2.303 {R}}\left(\frac{1}{{~T}_{2}}-\frac{1}{{~T}_{1}}\right)\)
\(\log \left(\frac{{K}_{700}}{{~K}_{600}}\right)=\frac{-{E}_{{a}}}{2.303 {R}}\left(\frac{1}{700}-\frac{1}{600}\right)\)
\(\log \left(\frac{6.36 \times 10^{-3}}{{~K}_{600}}\right)=\frac{+209 \times 1000}{2.303 \times 8.31}\left(\frac{100}{700 \times 600}\right)\)
\(\log \left(6.36 \times 10^{-3}\right)-\log {K}_{600}=2.6\)
\(\Rightarrow \log {K}_{600}=-2.19-2.6=-4.79\)
\(\Rightarrow {K}_{600}=10^{-4.79}\)
\(=1.62 \times 10^{-5}\)
\(=16.2 \times 10^{-6}\)
\(=x \times 10^{-6}\)
\(\Rightarrow {x}=16\)
$NO_{(g)} + Br_{2 (g)} $ $\rightleftharpoons$ $ NOBr_{2 (g)} , NOBr_{2 (g)} + NO_{(g)}\rightarrow 2 NOBr_{(g)}$ જો બીજી પ્રક્રિયાએ વેગનિર્ણાયક તબક્કો હોય તો પ્રક્રિયાનો ક્રમ $NO_{(g)} $ ના સંદર્ભમાં........ હશે.
$[R] (molar)$ |
$1.0$ |
$0.76$ |
$0.40$ |
$0.10$ |
$t (min.)$ |
$0.0$ |
$0.05$ |
$0.12$ |
$0.18$ |
તો પ્રક્રિયાનો ક્રમ $...$ થશે.