Q 1 :    

Higher yield of NO in

N2(g)+O2(g)2NO(g) can be obtained at  [ΔH of the reaction =+180.7 kJ mol-1]


(A) higher temperature
(B) lower temperature
(C) higher concentration of N2
(D) higher concentration of O2

Choose the correct answer from the options given below:                  [2025]

  • B, C, D only

     

  • A, C, D only

     

  • A, D only

     

  • B, C only

     

(2)

N2(g)+O2(g)2NO(g);ΔH=+180.7 kJ mol-1

As the given reaction is endothermic, hence, it will go to the product side on increasing temperature. Also, on adding reactants, the reaction goes towards the product side.

Hence, conditions given in A, C, and D increase the production of NO(g).



Q 2 :    

For the reaction in equilibrium

N2(g)+3H2(g)2NH3(g),  ΔH=-Q

Reaction is favoured in the forward direction by                      [2024]

  • use of catalyst

     

  • decreasing concentration of N2

     

  • low pressure, high temperature and high concentration of ammonia

     

  • high pressure, low temperature and higher concentration of H2

     

(4)

According to Le Chatelier’s principle:
(i) Raising the temperature shifts the equilibrium to the left and decreases the equilibrium concentration of ammonia. In other words, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction and thus a catalyst is used.
    
(ii) If H2 is added to the reaction mixture at equilibrium, then the equilibrium of the reaction is disturbed. In order to restore it, the reaction proceeds in a direction wherein H2 is consumed, i.e., more of H2 and N2 react to form NH3 and finally the equilibrium shifts in the right (forward) direction.
    
(iii) As the pressure increases, the equilibrium shifts in the forward direction, a direction in which the number of moles of the gas decreases.



Q 3 :    

Which one of the following conditions will favour maximum formation of the product in the reaction A2(g)+B2(g)X2(g),  ΔrH=-X kJ ?    [2018]

  • Low temperature and high pressure

     

  • Low temperature and low pressure

     

  • High temperature and high pressure

     

  • High temperature and low pressure

     

(1)

On increasing the pressure and decreasing the temperature, equilibrium will shift in forward direction.
 

 



Q 4 :    

For the reversible reaction,

N2(g)+3H2(g)2NH3(g)+heat

The equilibrium shifts in the forward direction                        [2014]

  • by increasing the concentration of NH3(g)

     

  • by decreasing the pressure

     

  • by decreasing the concentrations of N2(g) and H2(g)

     

  • by increasing pressure and decreasing temperature

     

(4)

As the forward reaction is exothermic and leads to lowering of pressure (produces lesser number of gaseous moles) hence, according to Le Chatelier’s principle, at high pressure and low temperature, the given reversible reaction will shift in forward direction to form more product.
 

 



Q 5 :    

For a given exothermic reaction, Kp and Kp' are the equilibrium constants at temperatures T1 and T2, respectively. Assuming that heat of reaction is constant in temperature range between T1 and T2, it is readily observed that           [2014]

  • Kp>Kp'

     

  • Kp<Kp'

     

  • Kp=Kp'

     

  • Kp=1Kp'

     

(1)


logK'pKp=-ΔH2.303R[1T2-1T1]

For exothermic reaction,ΔH=-ve, i.e., heat is evolved. The temperature T2 is higher than T1.


Thus, (1T2-1T1) is negative.

So, logK'p-logKp=-ve or logKp>logK'p

or Kp>K'p