In a reaction, $2X \to Y$ , the concentration of $X$ decreases from $0.50\, M$ to $0.38\, M$ in $10\, min$ . what is the rate of reaction in $Ms^{-1}$ during the interval ?
$2 \times 10^{-4}$
$4 \times 10^{-2}$
$2 \times 10^{-2}$
$1 \times 10^{-2}$
Rate constant of reaction is $1.388 \times 10^{-3}\, mole^{-2}\,lit^{-2}\,sec^{-1}$ order of reaction will be
For the elementary reaction $M \rightarrow N$, the rate of disappearance of $M$ increases by a factor of $8$ upon doubling the concentration of $M$. The order of the reaction with respect to $M$ is :
For a reaction $2A + B \to $ Products, doubling the initial concentration of both the reactants increases the rate by a factor of $8$, and doubling the concentration of $+B$ alone doubles the rate. The rate law for the reaction is
The rate constant for a second order reaction is $8 \times {10^{ - 5}}\,{M^{ - 1}}\,mi{n^{ - 1}}$. How long will it take a $ 1\,M $ solution to be reduced to $0.5\, M$
The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the following table:
Experiment | $[ A ] / mol\, ^{-1}$ | $[ B ] / mol\, ^{-1}$ | Initial rate $/$ $mol$ $L^{-1}$ $min$ $^{-1}$ |
$I$ | $0.1$ | $0.1$ | $2.0 \times 10^{-2}$ |
$II$ | - | $0.2$ | $4.0 \times 10^{-2}$ |
$III$ | $0.4$ | $0.4$ | - |
$IV$ | - | $0.2$ | $2.0 \times 10^{-2}$ |