Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
It is the enthalpy change taking place during the reaction when the number of moles of reactants and products are same as the stoichiometric coefficient indicates in the balanced chemical equation. The enthalpy change of the reaction depends upon the conditions like temperature, pressure etc under which the chemical reaction is carried out. Therefore, it is necessary to select the standard state conditions. According to thermodynamics conventions, the standard state refers to 1 bar pressure and 298 K temperature. The enthalpy change of a reaction at this standard state conditions is called standard enthalpy of the reaction.(ΔHo)
(i) Enthalpy of formation: Enthalpy change when one mole of a given compound is formed from its elements.
H2(g) + 1/2O2(g) ———> 2H2O(l), ΔH = –890.36 kJ / mol
Exercise:
Calculate for chloride ion from the following data:
1/2 H2 (g) + 1/2 Cl2 (g) ———> HCl (g) ΔH = –92.4 KJ
HCl (g) + nH2O ——> H+ (aq) + Cl– (aq) ΔH = –74.8 KJ
ΔH1o (H+(aq)) = 0.0 KJ
(ii) Enthalpy of combustion: Enthalpy change when one mole of a substance is burnt in oxygen.
CH4 + 2O2(g) ———> CO2 + 2H2O(l), ΔH = –890.36 kJ / mol
(iii) Enthalpy of Neutralization: Enthalpy change when one equivalent of an acid is neutralized by a base or vice – versa in dilute solution. This is constant and its value is –13.7 kcal for neutralization of any strong acid by a base since in dilute solutions they completely dissociate into ions.
H+ (aq) + OH– (aq) ——> H2O(l), ΔH = –13.7 kcal
For weak acids and bases, heat of neutralization is different because they are not dissociated completely and during dissociation some heat is absorbed. So total heat evolved during neutralization will be less.
e.g. HCN + NaOH ——> NaCN + H2O, ΔH = –2.9 kcal
Heat of ionization in this reaction is equal to (–2.9 + 13.7) kcal = 10.8 kcal
(A) H+ + OH-—> H2O
(B) H2O + H+ —> H3O+
(C) 2H2 + O2 = 2H2O
(D) CH3COOH+ NaOH = CH3COONa + H2O
Solution: (A) Since heat of neutralization of strong acid and strong base is equal to the heat of formation of water.
i.e., NaOH + HCl —> NaCl + H2O + Q
Were Q = heat of neutralization
=> Na+ + OH– + H+ + Cl– —> Na++Cl– + H2O + Q
=> H+ + OH– —> H2O + Q
(iv) Enthalpy of hydration: Enthalpy of hydration of a given anhydrous or partially hydrated salt is the enthalpy change when it combines with the requisite no.of mole of water to form a specific hydrate. For example, the hydration of anhydrous copper sulphate is represented by
CuSO4(s) + 5H2O (l) ——> CuSO45H2O(s), ΔH° = –18.69 kcal
SOLVED EXAMPLE: Ionisation energy of Al = 5137 kJ mole–1 (ΔH) hydration of Al3+ = – 4665 kJ mole–1. (DH)hydration for Cl– = – 381 kJ mole–1. Which of the following statement is correct
(A) AlCl3 would remain covalent in aqueous solution
(B) Only at infinite dilution AlCl3 undergoes ionisation
(C) In aqueous solution AlCl3 becomes ionic
(D) None of these
Solution: If AlCl3 is present in ionic state in aqueous solution, therefore it has Al3+ & 3Cl–ions
Standard heat of hydration of Al3+ & 3Cl- ions
= – 4665 + 3 × (–381) kJ mole–1 = -5808 kJ/mole
Required energy of ionisation of Al = 5137 kJ mole–1
∴ Hydration energy overcomes ionisation energy
∴ AlCl3 would be ionic in aqueous solution
Hence (C) is the correct answer.
(v) Enthalpy of Transition: Enthalpy change when one mole of a substance is transformed from one allotropic form to another allotropic form.
C (graphite) ——> C(diamond), ΔH° = 1.9 kJ/mol
Solved example: The heat of transition for carbon from the following is
CDiamond + O2(g) ——> CO2(g) ΔH = – 94.3 kcal
CAmorphous + O2(g) ——> CO2(g) ΔH = – 97.6 kcal
(A) 3.3 kJ / mol (B) 3.3 kcal / mol
(C) –3.3 kJ / mol (D) – 3.3 kcal / mol
Solution: Given
CD + O2(g) ——> CO2(g) ΔH = –94.3 kcal/mole …(1)
CA + O2(g) ——> CO2(g) ΔH = – 97.6 kcal/mole …(2)
———————————————————————————
Subtracting equation (2) from equation (1):
CD – CA ——> 0; ΔH = +3.3 kcal/mole
CD ——> CA ΔH = +3.3 kcal/mole
(B)
Solved example: From the reaction P(white) ——> P (Red): ΔH = - 18.4 kJ, It follows that
(A) Red P is readily formed from white P
(B) White P is readily formed from red P
(C) White P can not be converted to red P
(D) White P can be converted into red P and red P is more stable
Solution: (D)
To read more, Buy study materials of Thermodynamics comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
THIRD LAW OF THERMODYNAMICS:- In all heat engines,...
Enthalpy of a System The quantity U + PV is known...
Specific Heat Capacity or Specific Heat [c] Table...
HESS’S LAW This law states that the amount...
Level 2 Objective Problems of Thermodynamics Level...
Work Done During Isothermal Expansion Table of...
Introduction to Thermodynamics Table of Content...
Application of Hess's Law 1. Calculation of...
The First Law of Thermodynamics:- The first law of...
Objective Questions of Thermodynamics and Answers...
Work Done During Adiabatic Expansion Table of...
Solved Problems on Specific Heat, Latent Heat and...
Solved Problems on Thermodynamics:- Problem 1:- A...
Second Law of Thermodynamics Table of Content...
Specific Heat Capacity and Its Relation with...
Miscellaneous Exercises Thermal Physics:- Problem:...
Gibbs Free Energy This is another thermodynamic...
Reversible and Irreversible Process:- Reversible...
BOMB CALORIMETER The bomb calorimeter used for...
Application of bond energies (i) Determination of...
Relationship between free Energy and Equilibrium...
Solved Examples on Thermodynamics:- Problem 1 :- A...
Thermodynamic Change or Thermodynamic Process...
Macroscopic Properties He properties associated...
Thermodynamic State of a System and Macroscopic...