(i) Define the following:(a) Standard enthalpy of formation (∆fH0).(

1.	(i) Define the following:(a) Standard enthalpy of formation (∆fH0).(b) Standard enthalpy of combustion ∆cH0 .Or(i) Define the following (a) Hess's law of constant heat summation. (b) First law of thermodynamics.(ii) Find out whether it is possible to reduce MgO using carbon at 298 K. If not, at what temperature it becomes spontaneous. For reactionMgO(s) + C(s) → Mg(s) + CO(g)∆rH0 = 91.18 kJ/mol,∆rS0 = 197.67 JK-1 mol-1 .
Answer» (i) (a) Enthalpy change accompanying the formation of 1 mole of a compound from its constituent elements, all substances being in their standard states (1 bar pressure and 298 K). (b) It is the enthalpy change accompanying the complete combustion of one mole of a substance in excess of oxygen or air. or (i) (a) Hess's law of constant Heat Summation: According to Hess's law, the standard enthalpy of reaction is the sum of algebraic sum of the standard enthalpies of reaction into which the overall reaction may be splitted or divided. (b) According to First law of Thermodynamics, energy can neither be created nor be destroyed although it may be changed from one form to another. (ii) ∆G = ∆H - T∆S = 91.18 × 10³ − 298 × 197.67 = 91180 − 58905.66 = 32274.34 For reaction to be spontaneous, ΔG should be negative. Temperature should be greater than 298 K for reaction to be spontaneous.

(i) Define the following:(a) Standard enthalpy of formation (∆fH0).(b) Standard enthalpy of combustion ∆cH0 .Or(i) Define the following (a) Hess's law of constant heat summation. (b) First law of thermodynamics.(ii) Find out whether it is possible to reduce MgO using carbon at 298 K. If not, at what temperature it becomes spontaneous. For reactionMgO(s) + C(s) → Mg(s) + CO(g)∆rH0 = 91.18 kJ/mol,∆rS0 = 197.67 JK-1 mol-1 .

Answer»

(i) (a) Enthalpy change accompanying the formation of 1 mole of a compound from its constituent elements, all substances being in their standard states (1 bar pressure and 298 K).

(b) It is the enthalpy change accompanying the complete combustion of one mole of a substance in excess of oxygen or air.

(i) (a) Hess's law of constant Heat Summation:

According to Hess's law, the standard enthalpy of reaction is the sum of algebraic sum of the standard enthalpies of reaction into which the overall reaction may be splitted or divided.

(b) According to First law of Thermodynamics, energy can neither be created nor be destroyed although it may be changed from one form to another.

(ii) ∆G = ∆H - T∆S

= 91.18 × 10³ − 298 × 197.67

= 91180 − 58905.66 = 32274.34

For reaction to be spontaneous, ΔG should be negative. Temperature should be greater than 298 K for reaction to be spontaneous.

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