Which of the following reactions do you expect to have the smallest entropy chan

Question

Which of the following reactions do you expect to have the smallest entropy change? A. 2HF(g) + CI_2(g) rightarrow 2HCI(g) + F_2(g) B. 2Fe(s) + 3/2O_2(g) rightarrow Fe_2O_3(s) C. CH_2(g) + 2O_2(g) rightarrow 2H_2O(l)K. D. Cu(s) + 1/2O_2(g) rightarrow CuO(s) 8. Calculate the encropy change of the following reaction at 298K from the data in your text.CI_2(g) + H_2(g) rightarrow 2HCL(g) 19. The entropy of vaporization foe SBr_a is 80.5 JK^-1. The enthalpy of vaporization is 27.5 kJmol^-1. What is the normal boiling point for SBr_a? 26. For the following reaction at 25degreeC: I_2(g)+CI_2(g) 2ICI(g) Calculate DeltaGdegree for the reaction in kilojoules. DeltaHdegree = -26.9 kJ and DeltaSdegree = 11.3J/K 27. Calculate DeltaG_fdegree for the reaction: C(s) + CO_2(g) rightarrow 2CO(g) Given: DeltaG_fdegree; CO_2 = -394.4kJ/mol CO = -137.2 kJ/mol a. -120 kJ b. +120 kJ c. -257.2 kJ d. +257.2 kJ e. cannot determine because DeltaG_fdegree has not been given for C(s). 28 If a reaction is endothermic nonspontaneous at 25degreeC, then it A. can nevev be spontaneous. B can become spontaneous by adding a catalyst. C. may be spontaneous at lower temperature. D. may be spontaneous at lower temperature E. is euthoremic and spontaneous at high temperature.

Explanation / Answer

1. Reaction A : ( 2HF(g)+Cl2(g)----------> 2HCl(g) + F2(g) will have smallest entropy change because there is no phase and number of moles of reactants are equal to that of products

2.Entropy of chlorine = 223.8 J/mol.K and H2=130.68 j/mol.K and that of HCl =186.9

Entropy change= 2* entropy of HCl- ( entropy og Cl2+ entropy of H2)

=2*186.9- (223.8+130.68)=19.32 J/K

3.

Normal boiling point = enthalpy of vaporization/ entropy of vaporization= 27.5*1000/80.5 K=341.6 K

4.

delG=delH- TdelS= -26.9*1000-298.15*11.3 =-30269.1 J =-30.269 Kj

5.

delG= delG of products – delG of reactants= 2*-137.2+394.4=120 KJ ( B is correct)

6. gibbs free energy delG has to be -ve for a   process to be spontaneous

delG= delH- TdelS

delH is +Ve for endothermic process. and for delG to be -ve TdelS has to be +ve. With an increase in temperature, TdelS increase such that TdelS>delH. So at high enough temperature, the Gibbs free energy change is -ve

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