Hydrogenation of vegetable ghee at $25°\mathrm{C}$ reduces the pressure of ${\mathrm{H}}_2$ from $2 \mathrm{atm}$ to $1.2 \mathrm{atm}$ in $50 \mathrm{mins}.$ Calculate the rate of reaction in terms of change in pressure per minute.

For a gas reaction at $\mathrm{T} \left(\mathrm{K}\right)$, the rate is given by $-\frac{{\mathrm{dP}}_{\mathrm{A}}}{\mathrm{dt}}={\mathrm{k}}^{\text{'}}{\mathrm{P}}_{\mathrm{A}}^2 \mathrm{atm} {\mathrm{hr}}^{-1}$. If the rate equation is expressed as $-{\mathrm{r}}_{\mathrm{A}}=-\frac{1}{\mathrm{V}}\frac{{\mathrm{dn}}_{\mathrm{A}}}{\mathrm{dt}}={\mathrm{kC}}_{\mathrm{A}}^2$,$\mathrm{mol} {\mathrm{L}}^{-1} {\mathrm{hr}}^{-1}$, the rate constant $\mathrm{k}$ is given by

(where, $\mathrm{R}=$ ideal gas law constant, $\mathrm{cal} {\mathrm{g}}^{-1} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}$)

In the following reaction $\mathrm{xA}\to \mathrm{yB}$,

$\log \left[-\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}\right]=\log \left[\frac{\mathrm{dC}}{\mathrm{dt}}\right]+0.3$.

where '$-$' sign indicates rate of disappearance of the reactant. Thus, $\mathrm{x}:\mathrm{y}$ is

A reaction follows the given concentration $(\mathrm{C} \mathrm{in} \mathrm{Molarity})$ vs time graph. What will be the rate for this reaction at $20$ seconds?

The instantaneous rate of disappearance of the ${\mathrm{MnO}}_4^{-}$ ion in the following reaction is $4.56\times {10}^{-3} {\mathrm{Ms}}^{-1}$. Then, what will be the rate of appearance of ${\mathrm{I}}_2$?

Reaction: $2{\mathrm{MnO}}_4^{-}+10{\mathrm{I}}^{-}+10{\mathrm{H}}^{+}\to 2{\mathrm{Mn}}^{2+}+5{\mathrm{I}}_2+8{\mathrm{H}}_2\mathrm{O}$