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Change in volume of the system does not alter the number of moles in which of the following equilibria?

  • Option 1)

    N_{2(g)}+O_{2(g)}\rightleftharpoons 2NO_{(g)}\; \; \;

  • Option 2)

    PC1_{5(g)}\rightleftharpoons PC1_{3(g)}+C1_{2(g)}\; \; \;

  • Option 3)

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

  • Option 4)

    SO_{2}C1_{2(g)}\rightleftharpoons SO_{2(g)}+C1_{2(g)}

 

Answers (1)

best_answer

As we learnt in 

Relation between Kp and Kc -

K_{p}=K_{c}(RT)^{\bigtriangleup n}

while calculating the value of Kp , pressure should be expressed in bar.

1\:bar=10^{5}pa

- wherein

\bigtriangleup n = (number of moles of gaseous products) - (number of moles gaseous reaction)

 

 If\Delta n=0 then there is no effect of change in volume an equilibrium 

For reaction 

N_{2} \: _{(g)}+O_{2_\:{(g)}}\rightleftharpoons 2NO _{(g)}\:,\Delta n is zero

In this reaction the ratio of number of moles of reactants to products is same i.e. 2 : 2, hence change in volume will not alter the number of moles.

 


Option 1)

N_{2(g)}+O_{2(g)}\rightleftharpoons 2NO_{(g)}\; \; \;

This option is correct 

Option 2)

PC1_{5(g)}\rightleftharpoons PC1_{3(g)}+C1_{2(g)}\; \; \;

This option is incorrect 

Option 3)

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

This option is incorrect 

Option 4)

SO_{2}C1_{2(g)}\rightleftharpoons SO_{2(g)}+C1_{2(g)}

This option is incorrect 

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