The rate constant for the first order decomposition of \(H_2O _2\) is given by the following equation:
\(log\ k = 14.34 - 1.25 \times 10^4\ K/T \)
Calculate \(E_a\) for this reaction and at what temperature will its half-period be 256 minutes?
\(Arrhenius \ equation \ given\ by, \)
\(k = Ae^{\frac {-E_a}{RT}}\)
\(ln\ k = ln \ A - \frac {E_a}{RT}\)
\(ln\ k = log \ A - \frac {E_a}{RT}\)
⇒\(log\ k = log \ A - \frac {E_a}{2.303\ T}\) .....(i)
The given equation is
\(log\ k = 14.34-1.25×10^4\ K/T\) ......(ii)
From equation (i) and (ii), we obtain
\(\frac {E_a}{2.303 \ RT} = 1.25×10^4\ K/T\)
⇒ \(Ea=1.25×10^4K×2.303×R\)
= \(1.25 × 10^4K × 2.303 × 8.314 J K^{- 1}mol^ {- 1}\)
= \(239339.3 \ J mol^{-1} (approximately)\)
= \(239.34\ kJ mol^{-1}\)
Also, when \(t_{\frac 12}=256 \ minutes\),
\(k = \frac {0.693}{t_{1/2}}\)
\(k = \frac {0.693}{256}\)
\(k= 2.707×10^{-3} min^{-1}\)
\(k= 4.51 × 10^{-5} s^{-1}\)
It is also given that, \(log\ k= 14.34 - 1.25 × 10^4 K/T\)
⇒\(log(4.51×10^{-5})=14.34-1.25×10^4 K/T\)
\(log \ (0.654-05)=14.34-1.25×10^4 K/T\)
\(\frac {1.25×10^4K}{T} = 18.686\)
⇒\(T =\frac {1.25 \times 10^4 \ K}{18.686}\)
\(T = 668.95 \ K\)
\(T = 669\ K\ (approximately)\)
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
| Experiment | Time/s-1 | Total pressure/atm |
| 1 | 0 | 0.5 |
| 2 | 100 | 0.6 |
The rate constant for the decomposition of \(N_2O_5\) at various temperatures is given below:
| T/°C | 0 | 20 | 40 | 60 | 80 |
| 105 x k/s-1 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph between ln k and \(\frac 1T\) and calculate the values of \(A\) and \(E_a\).
Predict the rate constant at 30 ºC and 50 ºC.
The rate constant for the decomposition of hydrocarbons is 2.418 x 10-5 s-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Consider a certain reaction \(A\) \(→\) \(Products\) with \(k = 2.0 \times 10^{-2 }s^{-1}\) . Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L-1.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with \(t_{\frac 12} = 3.00\ hours\). What fraction of sample of sucrose remains after \(8 \ hours\)?
The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate equation of the reaction will given in the question.