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Engineering
91 Views   |  

An ideal gas undergoes a cyclic process as shown in Figure.

\Delta U_{BC}= -5 kJ mol^{-1}, q_{AB} = 2kJ mol^{-1}

W_{AB}= -5 kJ mol^{-1}, W_{CA} = 3kJ mol^{-1} 

Heat absorbed by the system during process CA is :

 

  • Option 1)

    -5kJ mol^{-1}

     

     

     

  • Option 2)

    +5kJ mol^{-1}

  • Option 3)

    18kJ mol^{-1}

  • Option 4)

    -18kJ mol^{-1}

 
As we learned   Isochoric Process - Volume is constant during the process - wherein     Isobaric process - Pressure is constant during the process - wherein      Isobaric  Isochoric  Not Defined               2 - 5  =  -3KJ                     -3 - 5 = -3KJ                         Q+w 8 = Q+3 Q= +5KJ     Option 1)       Option 2) Option 3) Option 4)
Engineering
106 Views   |  

For which of the following reactions, \Delta H is equal to \Delta U ?

  • Option 1)

    N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)

     

     

     

  • Option 2)

    2HI(g)\rightarrow H_{2}(g)+I_{2}(g)

  • Option 3)

    2NO_{2}(g)\rightarrow N_{2}O_{4}(g)

  • Option 4)

    2SO_{2}(g)+O_{2}(g)\rightarrow 2SO_{3}(g)

 
  As we learned Molar heat capacity for isobaric process C(p) -   - wherein or     Therefore ;  Option 1)       Option 2) Option 3) Option 4)
Engineering
116 Views   |  

The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is −3263.9 kJ mol−1 at 258 C; heat of combustion (in kJ mol−1) of benzene at constant pressure will be :

(R=8.314 JK−1 mol−1)

  • Option 1)

    −3267.6
     

  • Option 2)

    4152.6
     

  • Option 3)

    −452.46
     

  • Option 4)

    3260

 
As we learned Molar heat capacity for isobaric process C(p) -   - wherein or     For chemical reactions;          Option 1) −3267.6   Option 2) 4152.6   Option 3) −452.46   Option 4) 3260
Engineering
197 Views   |  

Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for an exothermic
reaction ?

  • Option 1)

    A and D

  • Option 2)

    A and B

  • Option 3)

    B and C

  • Option 4)

    C and D

 
As we learned @1380 For exothermic reaction slope will be positive & intercept may be positive or negative for a graph of so correct graph are A & B   Option 1) A and D Option 2) A and B Option 3) B and C Option 4) C and D
Engineering
130 Views   |  

Given: 

(i) \; 2Fe_{2}O_{3}_{(s)} \rightarrow 4Fe_{(s)} + 3O_{2(g)}\; ; \;\Delta\:_{r}G^{o} = + 1487.0\:kJ\;mol^{-1}

(ii) \; 2CO_{(g)} + O_{2(g)}\rightarrow 2CO_{2(g)}\; ; \;\Delta\:_{r}G^{o} = -514.4\:kJ\;mol^{-1}

Free energy change, \Delta\:_{r}G^{o} for the reaction   2Fe_{2}O_{3(s)} + 6CO_{g} \rightarrow 4Fe_{(s)} + 6 CO_{2(g)}  will be :

  • Option 1)

    −112.4 kJ mol −1

  • Option 2)

    −56.2 kJ mol −1

  • Option 3)

    −168.2 kJ mol −1

  • Option 4)

    −208.0 kJ mol −1

 

As we have learnt,

 

ΔG{reaction} -

\Delta G _{reaction }= \sum \Delta G_{product}- \sum \Delta G_{reactant}
 

- wherein

\sum \Delta G_{product}=Sum of   \Delta G of all product

\sum \Delta G_{reactant}=Sum of   \Delta G of all reactant

 

 \Delta G_{Reaction} = \Delta G_{1} + 3\times \Delta G_{2} \\*\Rightarrow \Delta G_{Reaction} = 1487 + (3\times -514.4) = -56.2\;kJ\:mol^{-1}

 


Option 1)

−112.4 kJ mol −1

Option 2)

−56.2 kJ mol −1

Option 3)

−168.2 kJ mol −1

Option 4)

−208.0 kJ mol −1

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Engineering
799 Views   |  

fG0 at 500 K for substance ‘S’ in liquid state and gaseous state are +100.7 kcal mol −1 and +103 kcal mol−1 , respectively. Vapour pressure of liquid ‘S’ at 500 K is approximately equal to :
(R=2 cal K−1 mol−1 )

  • Option 1)

    0.1 atm

  • Option 2)

    1 atm

  • Option 3)

    10 atm

  • Option 4)

    100 atm

 
As we have learnt,   Δ G of equilibrium -   - wherein At Equilibrum     and     Option 1) 0.1 atm Option 2) 1 atm Option 3) 10 atm Option 4) 100 atm
Engineering
82 Views   |  

For which of the following processes, ΔS  is negative ?

  • Option 1)

    H_{2}(g)\rightarrow 2H(g)

  • Option 2)

    N_{2}(g,1 atm)\rightarrow N_{2}(g,5atm)

  • Option 3)

    C(diamond)\rightarrow C(graphite)

  • Option 4)

    N_{2}(g,273 K)\rightarrow N_{2}(g, 300K)

 
As we have learned Change in entropy for ideal gas in terms of C(p) - - wherein Molar heat capacity at constant pressure       for isothermal process            Option 1) This is incorrect Option 2) This is correct Option 3) This is incorrect Option 4) This is incorrect
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