| S. No. | Chemical equation | Molecularity | Rate law | Order w.r.t. | |||||||||||||||||||||
| First reactant | Second reactant | Overall | |||||||||||||||||||||||
| 1. | \[aA+bB\to \]product | a + b | \[\left( \frac{dx}{dt} \right)=k{{[A]}^{a}}{{[B]}^{b}}\] | a | b | a + b | |||||||||||||||||||
| 2. | \[aA+bB\to \]product | a + b | \[\left( \frac{dx}{dt} \right)=k{{[A]}^{2}}{{[B]}^{0}}\] | 2 | zero, if B is in excess | 2 | |||||||||||||||||||
| 3. | \[2{{H}_{2}}{{O}_{2}}\xrightarrow{Pt,\Delta }2{{H}_{2}}O+{{O}_{2}}\] | 2 (Bimolecular) | \[\left( \frac{dx}{dt} \right)=k[{{H}_{2}}{{O}_{2}}]\] | 1* | ----- | 1 | |||||||||||||||||||
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4. more...
On the basis of reaction rates, the chemical reactions have been classified into the following three types,
(1) Very fast or instantaneous reactions : These reactions occur at a very fast rate generally these reactions involve ionic species and known as ionic reactions. It is almost impossible to determine the rates of these reactions.
Examples
(i) \[AgN{{O}_{3}}+NaCl\to \underset{\text{(PPt}\text{.)}}{\mathop{AgCl}}\,+NaN{{O}_{3}}\] (Precipitation reaction)
(ii) \[\underset{\text{(acid)}}{\mathop{HCl}}\,+\underset{\text{(base)}}{\mathop{NaOH}}\,\to \underset{\text{(Salt)}}{\mathop{NaCl}}\,+{{H}_{2}}O\] (Neutralization reaction)
(2) Moderate reaction : These reactions proceed with a measurable rates at normal temperature and it is these reactions are studied in chemical kinetics. Mostly these reactions are molecular in nature.
Examples
(i) Decomposition of \[{{H}_{2}}{{O}_{2}}\]: \[2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}\]
(ii) Decomposition of \[{{N}_{2}}{{O}_{5}}\]: \[2{{N}_{2}}{{O}_{5}}\to 2{{N}_{2}}{{O}_{4}}+{{O}_{2}}\]
(3) Very slow reactions : These reactions are extremely slow and take months together to show any measurable change.
Examples
(i) Rusting of iron : \[F{{e}_{2}}{{O}_{3}}+x{{H}_{2}}O\to \underset{\text{Hydrated ferric oxide (Rust)}}{\mathop{F{{e}_{2}}{{O}_{3}}.\,x{{H}_{2}}O}}\,\]
(ii) \[2{{H}_{2}}+{{O}_{2}}\xrightarrow{\text{Room temperature}}2{{H}_{2}}O\]
When a bond is formed between atoms, energy is released. Obviously same amount of energy will be required to break the bond. The energy required to break the bond is termed bond dissociation energy. The more precise definition is,
“The amount of energy required to break one mole of bond of a particular type between the atoms in the gaseous state, i.e., to separate the atoms in the gaseous state under 1 atmospheric pressure and the specified temperature is called bond dissociation energy.”
For example, \[H-H(g)\to 2H(g);\]\[\Delta H=+\,433\,kJ\,mo{{l}^{-1}}\]
\[Cl-Cl(g)\to 2Cl\,(g);\]\[\Delta H=+\,242.5\,kJ\,mo{{l}^{-1}}\]
\[H-Cl(g)\,\to H(g)+Cl(g);\]\[\Delta H=+\,431\,kJ\,mo{{l}^{-1}}\]
The bond dissociation energy of a diatomic molecule is also called bond energy. However, the bond dissociation energy depends upon the nature of bond and also the molecule in which the bond is present. When a molecule of a compound contains more than one bond of the same kind, the average value of the dissociation energies of a given bond is taken. This average bond dissociation energy required to break each bond in a compound is called bond energy.
Bond energy is also called, the heat of formation of the bond from gaseous atoms constituting the bond with reverse sign.
\[H(g)+Cl(g)\to H-Cl\,(g);\]\[\Delta H=-\,431\,kJ\,mo{{l}^{-1}}\]
Bond energy of \[H-Cl=-\](enthalpy of formation) \[=-(-431)=+\,431\,kJ\,mo{{l}^{-1}}\]
Consider the dissociation of water molecule which consists of two \[O-H\] bonds. The dissociation occurs in two stages.
\[{{H}_{2}}O(g)\to H(g)+OH(g);\,\]\[\Delta H=497.89\,kJ\,mo{{l}^{-1}}\]
\[OH(g)\to H(g)+O(g);\]\[\Delta H=428.5\,kJ\,mo{{l}^{-1}}\]
The average of these two bond dissociation energies gives the value of bond energy of \[O-H.\]
Bond energy of \[O-H\] bond
\[=\frac{497.8+428.5}{2}=463.15\,kJ\,mo{{l}^{-1}}\]
Similarly, the bond energy of \[N-H\] bond in \[N{{H}_{3}}\] is equal to one – third of the energy of dissociation of \[N{{H}_{3}}\] and those of \[C-H\] bond in \[C{{H}_{4}}\] is equal to one – fourth of the energy of dissociation of \[C{{H}_{4}}.\]
Bond energy of \[C-H=\frac{1664}{4}=416\,kJ\,mo{{l}^{-1}}\]
\[[C{{H}_{4}}(g)\to C(g)\,+4H(g);\]\[\Delta H=1664\,kJ\,mo{{l}^{-1}}]\]
Applications of bond energy
(1) Heat of a reaction \[=\Sigma \]Bond energy of reactants – \[\Sigma \] Bond energy of products.
(2) Determination of resonance energy : When a compound shows resonance, there is considerable difference between the heat of formation as calculated from bond energies and that determined experimentally.
Resonance energy = Experimental or actual heat of formation – Calculated heat of formation.
(1) Levoisier and Laplace law : According to this law enthalpy of decomposition of a compound is numerically equal to the enthalpy of formation of that compound with opposite sign, For example,
\[C(s)+{{O}_{2}}\to C{{O}_{2}}(g);\Delta H=-94.3\,kcal\]
\[C{{O}_{2}}(g)\to C(s)+{{O}_{2}}(g);\,\Delta H=+94.3kcal\]
(2) Hess's law (the law of constant heat summation) : This law was presented by Hess in 1840. According to this law “If a chemical reaction can be made to take place in a number of ways in one or in several steps, the total enthalpy change (total heat change) is always the same, i.e. the total enthalpy change is independent of intermediate steps involved in the change.” The enthalpy change of a chemical reaction depends on the initial and final stages only. Let a substance A be changed in three steps to D with enthalpy change from A to B, \[\Delta {{H}_{1}}\] calorie, from B to \[C,\,\Delta {{H}_{2}}\] calorie and from C to \[D,\,\Delta {{H}_{3}}\] calorie. Total enthalpy change from A to D will be equal to the sum of enthalpies involved in various steps. Total enthalpy change \[\Delta {{H}_{\text{steps}}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}+\Delta {{H}_{3}}\]
Now if D is directly converted into A, let the enthalpy change be \[\Delta {{H}_{\text{direct}}}.\] According to Hess's law \[\Delta {{H}_{\text{steps}}}+\Delta {{H}_{\text{direct}}}=0,\] i.e. \[\Delta {{H}_{\text{steps}}}\] must be equal to \[\Delta {{H}_{\text{direct}}}\] numerically but with opposite sign. In case it is not so, say \[\Delta {{H}_{\text{steps}}}\](which is negative) is more that \[\Delta {{H}_{\text{direct}}}\](which is positive), then in one cycle, some energy will be created which is not possible on the basis of first law of thermodynamics. Thus, \[\Delta {{H}_{\text{steps}}}\] must be equal to \[\Delta {{H}_{\text{direct}}}\] numerically.
(i) Experimental verification of Hess's law
(a) Formation of carbon dioxide from carbon
First method : carbon is directly converted into \[C{{O}_{2}}(g).\]
\[C(s)+{{O}_{2}}(g)=C{{O}_{2}}(g);\,\,\Delta H=-94.0\,kcal\]
Second method : Carbon is first converted into \[CO(g)\] and then \[CO(g)\] into \[C{{O}_{2}}(g)\], i.e. conversion has been carried in two steps,
\[C(s)+\frac{1}{2}{{O}_{2}}=CO(g)\] ; \[\Delta H=-26.0\,kcal\]
\[CO(g)+\frac{1}{2}{{O}_{2}}=C{{O}_{2}}(g);\] \[\Delta H=-\,68.0\,kcal\]
Total enthalpy change \[C(s)\] to \[C{{O}_{2}}(g);\] \[\Delta H=-94.0\,kcal\]
(b) Formation of ammonium chloride from ammonia and hydrochloric acid:
First method
\[\frac{\begin{align} & N{{H}_{3}}(g)+HCl=N{{H}_{4}}Cl(g);\Delta H=-\,42.2\,kcal\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & N{{H}_{4}}Cl\left( g \right)+aq=N{{H}_{4}}Cl\left( aq \right);DH=+4.0kcal \\ \end{align}}{~N{{H}_{3}}(g)+HCl(g)+aq=N{{H}_{4}}Cl(aq);\Delta H=-38.2\,kcal}\]
Second method
\[\frac{\begin{align} & N{{H}_{3}}(g)+aq=N{{H}_{3}}(aq);\Delta H=-8.4\,kcal\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & HCl(g)+aq=HCl(aq);\Delta H=-17.3\,kcal \\ & N{{H}_{3}}(aq)+HCl(aq)\,=N{{H}_{4}}Cl(aq);\Delta H=-12.3\,kcal \\ \end{align}}{N{{H}_{3}}(g)+HCl(g)+aq=N{{H}_{4}}Cl(aq);\Delta H=-38.0\,kcal}\]
(ii) Applications of Hess's law
(a) For the determination of enthalpies of formation of those compounds which cannot be prepared directly from the elements easily using enthalpies of combustion of compounds.
(b) For the determination of enthalpies of extremely slow reactions.
(c) For the determination of enthalpies of transformation of one allotropic form into another.
(d) For the determination of bond energies.
\[\Delta {{H}_{\text{reaction }}}=\Sigma \]Bond energies of reactants – \[\Sigma \]Bond energies of products.
(e) For the determination of resonance energy.
(f) For the determination of lattice energy.
Heat of reaction is defined as the amount of heat evolved or absorbed when quantities of the substances indicated by the chemical equation have completely reacted. The heat of reaction (or enthalpy of reaction) is actually the difference between the enthalpies of the products and the reactants when the quantities of the reactants indicated by the chemical equation have completely reacted. Mathematically,
Enthalpy of reaction (heat of reaction) \[=\Delta H=\Sigma {{H}_{P}}-\Sigma {{H}_{R}}\]
(1) Factors which influence the heat of reaction : There are a number of factors which affect the magnitude of heat of reaction.
(i) Physical state of reactants and products : Heat energy is involved for changing the physical state of a chemical substance. For example in the conversion of water into steam, heat is absorbed and heat is evolved when steam is condensed. Considering the following two reactions
\[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)={{H}_{2}}O(g);\] \[\Delta H=-\,57.8\,kcal\]
\[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)={{H}_{2}}O(l);\] \[\Delta H=-\,68.32\,kcal\]
It is observed that there is difference in the value of \[\Delta H\] if water is obtained in gaseous or liquid state. \[\Delta H\] value in second case is higher because heat is evolved when steam condenses. Hence, physical sate always affects the heat of reaction.
(ii) Allotropic forms of the element : Heat energy is also involved when one allotropic form of an element is converted into another. Thus, the value of \[\Delta H\] depends on the allotropic form used in the reaction. For example, the value of \[\Delta H\] is different when carbon in the form of diamond or in amorphous form is used.
C (diamond) \[+{{O}_{2}}(g)\,\to C{{O}_{2}}(g);\] \[\Delta H=-\,94.3\,kcal\]
C (amorphous) \[+{{O}_{2}}(g)\to C{{O}_{2}}(g);\] \[\Delta H=-\,97.6\,kcal\]
The difference between the two values is equal to the heat absorbed when 12g of diamond is converted into 12g of amorphous carbon. This is termed as heat of transition.
(iii) Temperature : Heat of reaction has been found to depend upon the temperature at which reaction is occurring. The variation of the heat of reaction with temperature can be ascertained by using Kirchhoff's equation.
\[\frac{\Delta {{H}_{{{T}_{2}}}}-\Delta {{H}_{{{T}_{1}}}}}{{{T}_{2}}-{{T}_{1}}}=\Delta {{C}_{P}}\]
Kirchhoff's equation at constant volume may be given as,
\[\frac{\Delta {{E}_{{{T}_{2}}}}-\Delta {{E}_{{{T}_{1}}}}}{{{T}_{2}}-{{T}_{1}}}=\Delta {{C}_{\nu }}\]
(iv) Reaction carried out at constant pressure or constant volume : When a chemical reaction occurs at constant volume, the heat change is called the internal energy of reaction at constant volume. However, most of the reactions are carried out at constant pressure; the enthalpy change is then termed as the enthalpy of reaction at constant pressure. The difference in the values is negligible when solids and liquids are involved in a chemical change. But, in reactions which involve gases, the difference in two values is considerable.
\[\Delta E+\Delta nRT=\Delta H\] or \[{{q}_{v}}+\Delta nRT={{q}_{p}}\]
\[\Delta E\] =\[{{q}_{v}}=\] heat change at constant volume;
\[\Delta E\] \[\Delta H\]=\[{{q}_{p}}=\] heat change at constant pressure,
\[\begin{align} & \Delta n=\text{total number of moles of gaseous product } \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{- total number of moles of gaseous reactants}\text{.} \\\end{align}\]
(2) Types of heat of reaction
(i) Heat of formation : It is the quantity of heat evolved or absorbed (i.e. more...
(1) Exothermic reactions : The chemical reactions which proceed with the evolution of heat energy are called exothermic reactions. The heat energy produced during the reactions is indicated by writing +q or more precisely by giving the actual numerical value on the products side. In general exothermic reactions may be represented as, \[A+B\to C+D+q\] (heat energy)
In the exothermic reactions the enthalpy of the products will be less than the enthalpy of the reactants, so that the enthalpy change is negative as shown below
\[\Delta H={{H}_{p}}-{{H}_{r}}\] ; \[{{H}_{p}}<{{H}_{r}}\]; \[\Delta H=-\,ve\]
Examples : (i) \[C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g)+393.5kJ\]
(at constant temperature and pressure)
or \[C(s)+{{O}_{2}}(g)\to C{{O}_{2}}(g);\] \[\Delta H=-393.5kJ\]
(ii) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(l);\] \[\Delta H=-285.8kJ\]
(iii) Fermentation is also an example of exothermic reaction.
(2) Endothermic reactions : The chemical reactions which proceed with the absorption of heat energy are called endothermic reactions. Since the heat is added to the reactants in these reactions, the heat absorbed is indicated by either putting (–) or by writing the actual numerical value of heat on the reactant side
\[A+B\to C+D-q\] (heat energy)
The heat absorbed at constant temperature and constant pressure measures enthalpy change. Because of the absorption of heat, the enthalpy of products will be more than the enthalpy of the reactants. Consequently, \[\Delta H\] will be positive \[(+ve)\] for the endothermic reactions.
\[\Delta H={{H}_{p}}-{{H}_{r}}\]; \[{{H}_{p}}>{{H}_{r}}\]; \[\,\,\Delta H=+ve\]
Example : (i) \[{{N}_{2}}(g)+{{O}_{2}}(g)\to 2NO(g);\,\Delta H=+180.5\,kJ\]
(ii) \[C(s)+2S(s)\to C{{S}_{2}}(l)\,\Delta H=+92.0kJ\]
(iii) Preparation of ozone by passing silent electric discharged through oxygen is the example of endothermic reaction.
(iv) Evaporation of water is also the example of endothermic reaction.
For exothermic reaction : \[\Delta H\] or \[\Delta E=-ve\]
For endothermic reaction : \[\Delta H\] or \[\Delta E=+ve\]
This law was first formulated by German chemist Walther Nernst in 1906. According to this law,
“The entropy of all perfectly crystalline solids is zero at the absolute zero temperature. Since entropy is a measure of disorder, it can be interpretated that at absolute zero, a perfectly crystalline solid has a perfect order of its constituent particles.”
The most important application of the third law of thermodynamics is that it helps in the calculation of absolute entropies of the substance at any temperature T.
\[S=2.303{{C}_{p}}\log T\]
Where CPis the heat capacity of the substance at constant pressure and is supposed to remain constant in the range of 0 to T.
Limitations of the law
(1) Glassy solids even at 0oK has entropy greater than zero.
(2) Solids having mixtures of isotopes do not have zero entropy at 0oK. For example, entropy of solid chlorine is not zero at 0oK.
(3) Crystals of CO, N2O, NO, H2O, etc. do not have perfect order even at 0oK thus their entropy is not equal to zero.
Gibb's free energy (G) is a state function and is a measure of maximum work done or useful work done from a reversible reaction at constant temperature and pressure.
(1) Characteristics of free energy
(i) The free energy of a system is the enthalpy of the system minus the product of absolute temperature and entropy i.e., \[G=H-TS\]
(ii) Like other state functions E, H and S, it is also expressed as \[\Delta G\]. Also \[\Delta G=\Delta H-T\Delta {{S}_{system}}\]where \[\Delta S\]is entropy change for system only. This is Gibb's Helmholtz equation.
(iii) At equilibrium \[\Delta G=0\]
(iv) For a spontaneous process decrease in free energy is noticed i.e., \[\Delta G=-ve\].
(v) At absolute zero, \[T\Delta S\]is zero. Therefore if \[\Delta G\]is – ve, \[\Delta H\]should be – ve or only exothermic reactions proceed spontaneously at absolute zero.
(vi) \[\Delta {{G}_{system}}=T\Delta {{S}_{universe}}\], where \[\Delta H=0\]
(vii) The standard free energy change,
\[\Delta {{G}^{o}}=-2.303RT{{\log }_{10}}\,K,\] where K is equilibrium constant.
(a) Thus if \[K>1,\]then \[\Delta {{G}^{o}}=-ve\]thus reactions with equilibrium constant K>1 are thermodynamically spontaneous.
(b) If K<1, then \[\Delta {{G}^{o}}=+ve\] and thus reactions with equilibrium constant K<1 are thermodynamically spontaneous in reverse direction.
(2) Criteria for spontaneity of reaction : For a spontaneous change \[\Delta G=-ve\] and therefore use of \[\Delta G=\Delta H-T\Delta S,\]provides the following conditions for a change to be spontaneous.
Criteria for spontaneity of reaction
(1) Definition : Entropy is a thermodynamic state quantity which is a measure of randomness or disorder of the molecules of the system.
Entropy is represented by the symbol “S”. It is difficult to define the actual entropy of a system. It is more convenient to define the change of entropy during a change of state.
The entropy change of a system may be defined as the integral of all the terms involving heat exchanged (q) divided by the absolute temperature (T) during each infinitesimally small change of the process carried out reversibly at constant temperature.
\[\Delta S={{S}_{final}}-{{S}_{initial}}=\frac{{{q}_{rev}}}{T}\]
If heat is absorbed, then \[\Delta S=+ve\]and if heat is evolved, then \[\Delta S=-ve.\]
(2) Units of entropy : Since entropy change is expressed by a heat term divided by temperature, it is expressed in terms of calorie per degree, i.e.,cal deg-1 . In SI units, the entropy is expressed in terms of joule per degree Kelvin, i.e., \[J{{K}^{-1}}\].
(3) Characteristics of entropy : The important characteristics of entropy are summed up below
(i) Entropy is an extensive property. Its value depends upon the amount of the substance present in the system.
(ii) Entropy of a system is a state function. It depends upon the state variables \[(T,p,V,n)\].
(iii) The change in entropy in going from one state to another is independent of the path.
(iv) The change in entropy for a cyclic process is always zero.
(v) The total entropy change of an isolated system is equal to the entropy change of system and entropy change of the surroundings. The sum is called entropy change of universe.
\[\Delta {{S}_{\text{universe}}}=\Delta {{S}_{sys}}+\Delta {{S}_{Surr}}\]
(a) In a reversible process, \[\Delta {{S}_{universe}}=0\]and, therefore
\[\Delta {{S}_{sys}}=-\Delta {{S}_{Surr}}\]
(b) In an irreversible process, \[\Delta {{S}_{universe}}>0\]. This means that there is increase in entropy of universe is spontaneous changes.
(vi) Entropy is a measure of unavailable energy for useful work.
Unavailable energy = Entropy × Temperature
(vii) Entropy, S is related to thermodynamic probability (W) by the relation,
\[S=k\,{{\log }_{e}}W\,\,\text{and}\,\,S=\text{ 2}\text{.303}k\text{ lo}{{\text{g}}_{\text{10}}}W\]
where, k is Boltzmann's constant
(4) Entropy changes in system & surroundings and total entropy change for Exothermic and Endothermic reactions : Heat increases the thermal motion of the atoms or molecules and increases their disorder and hence their entropy. In case of an exothermic process, the heat escapes into the surroundings and therefore, entropy of the surroundings increases on the other hand in case of endothermic process, the heat enters the system from the surroundings and therefore. The entropy of the surroundings decreases.
In general, there will be an overall increase of the total entropy (or disorder) whenever the disorder of the surroundings is greater than the decrease in disorder of the system. The process will be spontaneous only when the total entropy increases.
(5) Entropy change during phase transition : The change of matter from one state (solid, liquid or gas) to another is called phase transition. Such changes occur at definite temperature such as melting point (solid to liquid). boiling point (liquid to more...
Carnot, a French engineer, in 1824 employed merely theoretical and an imaginary reversible cycle known as carnot cycle to demonstrate the maximum convertibility of heat into work.
The system consists of one mole of an ideal gas enclosed in a cylinder fitted with a piston, which is subjected to a series of four successive operations.
For cyclic process, the essential condition is that net work done is equal to heat absorbed. This condition is satisfied in a carnot cycle.
\[\frac{w}{{{q}_{2}}}=\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}}=\]Thermodynamic efficiency
Thus, the larger the temperature difference between high and low temperature reservoirs, the more the heat converted into work by the heat engine.
Since \[\frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}}<1\], it follows that \[w<{{q}_{2}}\]. This means that only a part of heat absorbed by the system at the higher temperature is transformed into work. The rest of the heat is given out to surroundings. The efficiency of the heat engine is always less then 1. This has led to the following enunciation of the second law of thermodynamics.
It is impossible to convert heat into work without compensation.
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