From the following which is a second order reaction
$K = 5.47 \times {10^{ - 4}}\,{\sec ^{ - 1}}$
$K = 3.9 \times {10^{ - 3}}\,{\rm{mole}}\,\,{\rm{lit}}\,\,{\sec ^{ - 1}}$
$K = 3.94 \times {10^{ - 4}}\,{\rm{lit}}\,\,{\rm{mol}}{{\rm{e}}^{{\rm{ - 1}}}}\,\,{\sec ^{ - 1}}$
$K = 3.98 \times {10^{ - 5}}{\rm{lit}}\,\,{\rm{mol}}{{\rm{e}}^{{\rm{ - 2}}}}\,\,{\sec ^{ - 1}}$
The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by graph
For the following rate law determine the unit of rate constant. Rate $=-\frac{d[ R ]}{d t}=k[ A ]^{\frac{1}{2}}[ B ]^{2}$
For the reaction taking place on water, the order of reaction is
${{H}_{2}}+C{{l}_{2}}\xrightarrow{\text{Sunlight}}2HCl$
The rate of certain reaction depends on concentration according to the equation $\frac{{ - dc}}{{dt}}\, = \,\frac{{{K_1}C}}{{1 + {K_2}C}},$ what is the order, when concentration $(c)$ is very-very high
The rate of a gaseous reaction is given by the expression $K\,[A]\,[B]$. If the volume of the reaction vessel is suddenly reduced to $1/4^{th} $ of the initial volume, the reaction rate relating to original rate will be