The rate of dissappearance of $MnO_4^-$ in the following reaction is $4.56 \times 10^{-3}\, M/s$

$2MnO_4^-+ 10I^-+ 16H^+ \to 2Mn^{2+} + 5I_2 + 8 H_2O$

The rate of apperance of $I_2$ is

  • A

    $1.14 \times {10^{ - 3}}\,M/s$

  • B

    $5.7 \times {10^{ - 3}}\,M/s$

  • C

    $4.56 \times {10^{ - 4}}\,M/s$

  • D

    $1.14 \times {10^{ - 2}}\,M/s$

Similar Questions

Consider the data given below for hypothetical reaction $A \to X$

$Time  (sec)$                     Rate $(mol\,  L^{-1} sec.^{-1})$

$0$                                      $1.60 \times 10^{-2}$

$10$                                    $1.60 \times 10^{-2}$

$20$                                    $1.60 \times 10^{-2}$

$30$                                    $1.60 \times 10^{-2}$

From the above data, the order of reaction is

The reaction $CH _{3} COF + H _{2} O \quad \rightleftharpoons CH _{3} COOH + HF$

Condition $I$ $:$ $\left[ H _{2} O \right]_{0}=1.00 \,M$

                           $\left[ CH _{3} COF \right]_{0}=0.01 \,M$

Condition $II$ $:$ $\left[ H _{2} O \right]_{0}=0.02 \,M$

                             $\left[ CH _{3} COF \right]_{0}=0.80 \,M$

Condition - $I$ Condition - $II$

Time
$(t)$

$min$

$\left[ CH _{3} COF \right]$ $M$

Time
$(t)$

$min$

$\left[ H _{2} O \right] \,M$
$0$ $0.01000$ $0$ $0.0200$
$10$ $0.00867$ $10$ $0.0176$
$20$ $0.00735$ $20$ $0.0156$
$40$ $0.00540$ $40$ $0.0122$

Determine the order of reaction and calculate rate constant.

The half life of forward and reverse reactions are $400\,\,sec$ and $100\,\,sec$ and these half lives are independent from the concentration of Reactant then find the equilibrium constant of the reaction

Write unit of rate constant of following reaction :

$1.$ $\frac {1}{2}$ order

$2.$ $\frac {3}{2}$ order

Half-life period of a first order reaction is $1386$ seconds. The specific rate constant of the reaction is

  • [AIPMT 2009]