CBSE Guess > Papers > Important Questions > Class XII > 2007 > Chemistry > Themodoynamics

Themodoynamics

Q. 1. 2Mg +O2→2MgO  ∆rSo  = -217JK-1 mol -1  ∆rHo= -1202KJ mol -1 Determine ∆ So surrounding and ∆Sototal .  Is this reaction feasible at 298 K?

Q. 2. Predict the sign of entropy  (a)Liquid crystallizes in to a solid (b)Temperature of a crystalline solid raised from 0k to100k (c)2NaHCO3→Na2CO3 +H2O +CO2  (d)H2(g)→2H(g)  (e)N2  (1atm)→N2(0.5atm)

Q. 3. The enthalpy of vaporization of benzene is 30.8KJ/mole at its boiling point of 80oC. Find ∆S (i) liquid to vapour (ii) vapour to liquid .

Q. 4. Calculate ∆r Soat 298K  2NH3 + C02→NH2CONH2+H2O  So NH2CONH2 =174JK-1Mol -1 , So H2O=70 JK-1Mol -1 So NH3 =192 JK-1Mol -1 So C02=214 JK-1Mol -1
     
Q. 5. Calculate ∆Gofor  the reaction C(graphite)+2H2(g)→CH4(g)  ∆H0CH4 =- -74.81KJ/mole S0 JK-1Mol -1  C{graphite} =5.8, H2=131,  CH4(g)=-−74.81

Q. 6. CaCO3 (s)→CaO(s)+CO2 (g) at1273k and 1bar pressure. Is this reaction spontaneous at 1273K?Calculate Kp at
            1273K. Also calculate partial pressure of CO2 at equilibrium.
                                           CaCO3(s)     CO2(g)    CaO(s)
                    ∆HoKJ/mole    -1206.9         -393.51    -635
                   
                   SoJK−mol−1          92.9                      214           39.75

Q. 7.  SiH4(g))+2O2(g)→SiO2(s) + 2H2O(g)      ∆Gof KJ/mole   SiH4(g ) =52.3   SiO2(s)= -805  H2O(g) = −228  Is this 
           reaction  spontaneous?

Q. 8.  Is it possible to reduce Fe2O3 using CO    
           Given   2 Fe2O3 (s)→4Fe(s) + 3O2(g)   ∆Go = 1487KJ/mole      2CO(g) + O2→2CO2(g)      ∆Go = −515KJ/mole

Q. 9. Calculate ∆Go     3/2 O2(g)→O3(g)     Kp=2.47×10−29

Q. 10. In a fuel cell  CH3OH(l)+ 3/2 O2(g→ CO2 (g+2H2O(l)  Calculate ∆Go for the reaction and the  efficiency of the cell
∆Go     for the combustion of  methanol is −726KJ/mole

Q. 11. 2NaHCO3(s) →Na2CO3(s)+H2O(g)+ CO2(g)     
   
∆Gof        CO2(g)      H2O(g)      Na2CO3(s)    NaHCO3(s)
KJ/mole   −395        −237         −1047            −851
      
      ∆Hof         −393        −285         −1131             −948
      KJ/mole      Calculate the  temperature above which  the reaction become  feasible

Q. 12.   CH3COOH+C2H5OH→CH3COOC2H5  +H2O   Starting with 1mole each of the reactants,⅔ mole of each reactants
         consumed to attain equilibrium at 298K. Find ∆Goat 298K

Q. 13.   N2 (g)+3H2(g)→2NH3(g)∆rH=−95.4 KJ/mole  ∆rS=−198.3JK−1mol−1  (a) Determine the temperature at which ∆G=0
      Predict the nature of this reaction at this temperature,above this temperature and below this temperature.