The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by graph
$b$
$c$
$d$
$a$
For the non - stoichimetre reaction $2A + B \rightarrow C + D,$ the following kinetic data were obtained in three separate experiments, all at $298\, K.$
Initial Concentration $(A)$ |
Initial Concentration $(A)$ |
Initial rate of formation of $C$ $(mol\,L^{-1}\,s^{-1})$ |
$0.1\,M$ | $0.1\,M$ | $1.2\times 10^{-3}$ |
$0.1\,M$ | $0.2\,M$ | $1.2\times 10^{-3}$ |
$0.2\,M$ | $0.1\,M$ | $2.4 \times 10^{-3}$ |
The rate law for the formation of $C$ is :
The rate of reaction, $A + B + C \longrightarrow P$ is given by
$r = \frac{{ - d\left[ A \right]}}{{dt}} = K\,{\left[ A \right]^{\frac{1}{2}}}\,{\left[ B \right]^{\frac{1}{2}}}\,{\left[ C \right]^{\frac{1}{4}}}$
The order of reaction is
The reaction $2FeC{l_3} + SnC{l_2} \to 2FeC{l_2} + SnC{l_4}$ is an example of
The reaction $2NO + Br_2 \rightarrow 2NOBr,$ follows the mechanism given below
$(I)$ $NO + Br_2 \rightleftharpoons NOBr_2 $ ........ Fast
$(II)$ $NOBr_2 + NO \rightarrow 2NOBr$ ......... Slow
The overall order of this reaction is
If a reaction has the experimental rate expression rate $= K [A]^2[B]$, if the concentration of $A$ is doubled and the concentration of $B$ is halved, the what happens to the reaction rate