3) Consider the following chemical reaction that describes the formation of H20(

Question

3) Consider the following chemical reaction that describes the formation of H20(g) H2 (g)-yo(g)H2O(g) a. At 298 K, 11 for the reaction above is-241.8 kJ/mol and AS is-44.42 J/ mol-K. Calculate AH , AS , and G for the above reaction at 398 K. The constant pressure heat capacity of H2 (g) is 28.8 J/mol.K, O2 (g) is 29.4 J/mol-K, and H,0(g) is 33.6 J/mol-K -1 b· AS for this reaction at room temperature is less than zero. Why is this the case? Despite AS being negative for this reaction at room temperature, it is spontaneous at this temperature. How is this consistent with the second law of thermodynamics? c.

Explanation / Answer

entropy change= ln(T2/T1){1*CpH2O-(1*CPH2+0.5*CPO2)

T2= 398K, T1= 298K, 1,1 and 0.5 are coefficients of H2O, H2 and O2 respectively in the reaction

entropy change= ln( 398/273) {33.6-(28.8+29.4)}=-9.3 J/K

enthalpy change= deltaH298+ {1*CpH2O-(1*CPH2+0.5*CPO2) *(T2-T1)= -241.8*1000-24.6*(398-273)= -238725 J=238.725 Kj

deltaG= deltaH-T*deltaS= -238725-398*(-9.3)= -235024 J =-235.024 KJ

the reaction is spontaneous since deltaG is -ve.

since deltaS= sum of entropy of products- sum of entropy of reactants, since moles of reactants are more than moles of products, entropy change is -ve.

3. A reaction is spontaneous if deltaG <0 and deltaG= deltaH-T*deltaS

since deltaH is much more significant compared to deltaS for the reaction and at room temperature also, deltaG<0.

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