Question

Consider the spontaneous combustion reaction of hydrogen and oxygen to form water 2H_2(g) + o3(g) rightarrow 2H_2O(l) Delta H system = -483kJ Predicting the entropy change for the system. Delta S system from the balanced equation:

Answer 1

Consider the given reaction 2H2 (g)+02 (g)22H,0(1) AHsytem--483 kJ Part In the above reaction, the gaseous phase of reactant changes into liquid phase. The entropy of the system is decreasing because the entropy of liquid is less than the entropy of the gas Hence, the value of ASsaem will be negative Part 2 Since, the entropy change is not a driving force for the reaction and the enthalpy change is negative for the given reaction Hence, this entropy change is not a driving force for spontaneity Part 3 Since, the sign of ΔΡ,stem is negative for the given reaction. Hence, the given process is exothermic reaction Part 4 Since, the entropy of the given reaction is negative. Hence, the temperature of the surrounding will increase Part 5 Since, the temperature of surrounding is increases therefore, the entropy of surrounding is positive

Part 6 Since, the enthalpy is negative and exothermic reaction takes place therefore, the product is more stable than the reactant. Due to this reason it is driving force for spontaneity art Consider the reaction 2H2 (g)+ O2 (g)→2H2O(1) The value of entropy from entropy table is given below AS (H) = 130.6 J/mole K AS (O2) 205 J/mole K AS (H2O) = 69.9 J/mole K Calculate the AS for system, = (2x 69.9)-(2×130.6 +205) = 139.8-466.2 AS326.4 J/mole K Now, calculate AS for surrounding AH -483 x 1000J 298 K = 1620.81 J/K So, from above value of AS for surrounding is greater than AS for system in magnitude