ki/mol. G AG Look at the following reaction AG 70.91 k/mo. The forward reaction
ID: 499974 • Letter: K
Question
ki/mol. G AG Look at the following reaction AG 70.91 k/mo. The forward reaction is AG 141.8 /mol. The forward reaction is Suppose that you are given the following information for this react on at room AG" 141.8 kJ/mol. The reaction is at equilibrium. temperature 4G r for soa@ 300, 19 J/mol 4G r for sos(g) -371.1 klim ol Find the value for AG for this reaction and state whether the forward reaction is spontaneous, non spontaneous, or at equilibrium. 40 QID: 3771 o 201.4 J/mol K. Calculate the standard entropy change 471.5 J/mol. K at 25°C for the formation reaction shown 201.4 J/mol. K below. -132.7 J/mol K Use the following information: so 64.81 Jlmol for Pb(s) at 25°C .K So 205.3 JMnrol .K for O2Cg) at 25°C so 68.7 Jim ol .K for PbO2(s) at 25°C 5 QID: 3775 189.6 J/mol K. The value for AG for the following 3.3 x 105 J/mol reaction at 259C is -345.0 kJlmol. -4.1 x 105 J/mol e 4,7 x 105 J/mol Suppose that you raise the temperature to 93°C. What is the value for AGO for the reaction at C? (AH° -4.0 x 105 J/mol and As and -189.6 J/mol K)Explanation / Answer
for the reaction Pb(s) + O2(g)------->PbO2(s), entropy change= sum of entropy of products- sum of entropy of reactants
=1*68.7-{1* 205.3+1*64.81}=-201.41 J/K ( A is correct) ( where 1, 1 and 1 are coefficients of Pb, O2 and PbO2 respectively.
2. deltaG= deltaH-TdeltaS= -4*105+ (93+273)*189.6=-330606 J/K=-3.3*105 J/K( b is correct)
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