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
$1.14 \times {10^{ - 3}}\,M/s$
$5.7 \times {10^{ - 3}}\,M/s$
$4.56 \times {10^{ - 4}}\,M/s$
$1.14 \times {10^{ - 2}}\,M/s$
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 $min$ |
$\left[ CH _{3} COF \right]$ $M$ |
Time $min$ |
$\left[ H _{2} O \right] \,M$ |
$0$ | $0.01000$ | $0$ | $0.0200$ |
$10$ | $0.00867$ | $10$ | $0.0176$ |
$20$ | $0.00735$ | $20$ | $0.0156$ |
$40$ | $0.00540$ | $40$ | $0.0122$ |
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