Tag: rate of chemical reaction

The hydrolysis of ethyl acetate in an acidic medium is a:

The reaction $2N _2O _5(g)\, \rightarrow\, 4NO _2(g)\, +\, O _2(g)$ is first order w.r.t. $N _2O _5$. Which of the following graphs would yield a straight line ?

When ethyl acetate was hydrolyzed in the presence of $0.1 M$ $HCl$, the constant was found to be $5.40\, \times\, 10^{-5}\, s^{-1}$. But when $0.1$ $M\, H _2SO _4$ was used for hydrolysis, the rate constant was found to be $6.20\, \times\,10^{-5}\, s^{-1}$. From these we can say that:

For the reaction $2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$, if $\displaystyle -\, \frac{d[N _2O _5]}{dt}\, =\, k _1[N _2O _5]$, $\displaystyle \frac{d[NO _2]}{dt}\, =\, k _2[N _2O _5]$, $\displaystyle \frac{d[O _2]}{dt}\, =\, k _3[N _2O _5]$.
What is the relation between $k _1, k _2$ and $k _3$?

The rate law for the reaction : $:Ester+H^+\rightarrow Acid+Alcohol\,$ is
$V\,=\,k\;\left[ester \right]\;\left[H _3O^+ \right]^0$
What would be the new rate if
(a)$\;$conc. of ester is doubled
(b)$\;$conc. of $:H^{+}$ is doubled

In the presence of acid, the initial concentration of cane-sugar was reduced from 0.2 M to 0.1 in 5 hr and to 0.05 M in 10 hr. The reaction must be of :

The decomposition of $H _2O _2$ can be followed by titration with $KMnO _4$ and is found to be a first order reaction. The rate constant is $4.5\, \times\, 10^{-2}$. In an experiment, the initial titrate value was 25 mL. The titrate value will be 5 mL after a lapse of :

The half-life of decomposition of $N _2O _5$ is a first order reaction represented by:

The reaction $N _{2}O _{5}$ (in $CCl _{4}$) $\rightarrow 2NO _{2}+1/2O _{2}(g)$ is the first order in $N _{2}O _{5}$ with rate constant $6.2\times 10^{-4}S^{-1}$. 

The half life of decomposition of $N _2O _5$ is a first order reaction represented by
$N _2O _5\, \rightarrow\, N _2O _4\, =\, 1/2O _2$
After 15 min the volume of $O _2$ produced is $9mL$ and at the end of the reaction $35 mL$. The rate constant is equal to :