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.
$(i)$ Reaction is pseudo $1^{\text {st }}$ reaction, in which $\left[\mathrm{H}_{2} \mathrm{O}\right]$ is constant.
$(ii)$ Reaction is pseudo $1^{\text {st }}$ reaction, but in order to this reaction $=2$.
The rate of the reaction :
$2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.
$\frac{-d[N_2O_5 ]}{dt} = k[N_2O_5]$
$\frac{d[NO_2 ]}{dt} = k'[N_2O_5]\,;$ $\frac{d[O_2 ]}{dt} = k"[N_2O_5]$
The relationship between $k$ and $k'$ and betweenk and $k''$ are
The reaction $2FeC{l_3} + SnC{l_2} \to 2FeC{l_2} + SnC{l_4}$ is an example of
In a reaction $2A + B \to {A_2}B$, the reactant $ A $ will disappear at
When a reaction is progressing
In a reaction between $A$ and $B$, the initial rate of reaction $\left(r_{0}\right)$ was measured for different initial concentrations of $A$ and $B$ as given below:
$A/mol\,\,{L^{ - 1}}$ | $0.20$ | $0.20$ | $0.40$ |
$B/mol\,\,{L^{ - 1}}$ | $0.30$ | $0.10$ | $0.05$ |
${r_0}/mol\,\,{L^{ - 1}}\,\,{s^{ - 1}}$ | $5.07 \times 10^{-5}$ | $5.07 \times 10^{-5}$ | $1.43 \times 10^{-4}$ |
What is the order of the reaction with respect to $A$ and $B$?