Acid catalysed hydrolysis of ethyl acetate follows a pseudo-first order kinetics with respect to ester. If the reaction is carried out with large excess of ester, the order with respect to ester will be

• 1.5

• 0

• 0.5

• 1

The half-life for decay of ¹⁴C by β-emission is 5730 yr. The fraction of ¹⁴C decays, in a sample that is 22920 yr old, would be

• $\frac{1}{8}$

• $\frac{1}{16}$

• $\frac{7}{8}$

• $\frac{15}{16}$

A piece of wood from an archaeological sample has 5.0 counts min^-1 per gram of C-14, while a fresh sample of wood has a count of 15.0 min^-1g^-1. If half-life of C-14 is 5770 yr, the age of the archaeological sample is

• 8,500 yr

• 9,200 yr

• 10,000 yr

• 11,000 yr

Consider the following reaction for 2NO₂(g) + F₂(g) → 2NO₂Fg). The expression for the rate of reaction in terms of the rate of change of partial pressure of reactant and product is/are

• rate = $-\frac{1}{2}\left[\frac{\mathrm{dp}\left({\mathrm{NO}}_2\right)}{\mathrm{dt}}\right]$

• rate = $\frac{1}{2}\left[\frac{\mathrm{dp}\left({\mathrm{NO}}_2\right)}{\mathrm{dt}}\right]$

• rate = $-\frac{1}{2}\left[\frac{\mathrm{dp}\left({\mathrm{NO}}_2\mathrm{F}\right)}{\mathrm{dt}}\right]$

• rate = $\frac{1}{2}\left[\frac{\mathrm{dp}\left({\mathrm{NO}}_2\mathrm{F}\right)}{\mathrm{dt}}\right]$

The bacterial growth follows the rate law, $\frac{\mathrm{dN}}{\mathrm{dt}}$ = kN where k is a constant and 'N' is the dt number of bacteria cell at any time. If the population of bacteria (number of cell) is doubled in 5 min find the time in which the population will be eight times of the initial one?

$\mathrm{For} \mathrm{the} \mathrm{reaction}, 2{\mathrm{NO}}_2+ {\mathrm{F}}_2\to 2{\mathrm{NO}}_2\mathrm{F},$ following mechanism has been provided. 

$$\begin{gathered} {\mathrm{NO}}_2+ {\mathrm{F}}_2 \stackrel{\mathrm{slow}}{\to } {\mathrm{NO}}_2\mathrm{F} + \mathrm{F} \\ {\mathrm{NO}}_2+ {\mathrm{F}}_2 \stackrel{\mathrm{Fast}}{\to } {\mathrm{NO}}_2\mathrm{F} \end{gathered}$$

Thus, rate expression of the above reaction can be written as

• r =k [NO₂]²[F₂]

• r= k [NO₂] [F₂]

• r= k [NO₂]

• r= k [F₂]

For a reaction,  ${\mathrm{I}}^{-}+ {\mathrm{OCl}}^{-} \to {\mathrm{IO}}^{-} + \mathrm{Cl}$ an aqueous medium, the rate of reaction is given by:

                     $\frac{\mathrm{d}\left[{\mathrm{IO}}^{-}\right]}{\mathrm{dt}}= \mathrm{k} \frac{\left[{\mathrm{I}}^{-}\right] \left[{\mathrm{OCl}}^{-}\right]}{\left[\mathrm{OH}\right]}$

The overall order of reaction is

• -1

• 0

• 1

• 2

Reaction between compounds (A) and (B) occurs as follows

A (g) + 2B (g) → 2C (g)

Following results were obtained

The correct rate- law for the said reaction is (r and k rate and rate constant respectively)

• r = k[A]²B

• r = k[A][B]²

• r = k[A][B]

• r = k[A]²[B]²

In the reaction 2NO₂ $\underset{{\mathrm{k}}_2}{\overset{{\mathrm{k}}_1}{\rightleftharpoons }}$ N₂O₄; the rate of disappearance of NO₂ is equal to

• 2$\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}$ [NO₂]²

• 2k₁[NO₂]² - 2k₂[N₂O₄]

• 2k₂[NO₂]² - k₂[N₂O₄]

• (2k₁ - k₂)[NO₂]

The rate constant for the raection 2N₂O₅ → 4NO₂ + O₂ is 3.0 × 10^-5 s^-1. If rate is 2.40 × 10^-5, then concentration of N₂O₅ (in mol/L) is

• 1.4

• 1.2

• 0.04

• 0.8