The value of K_{P}/K_{C}   for the following reactions at 300 K are , respectively : (At 300 K , RT = 24.62 dm^{3}atm \: mol^{-1})

N_{2}(g)+ O_{2}(g)\rightleftharpoons 2NO(g)

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

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

 

  • Option 1)

    1,24.62 \: dm^{3}atm\: mol^{-1},

    606.0\: dm^{6}atm^{2}\: mol^{-2}

  • Option 2)

    1,24.62\: dm^{3}atm\: mol^{-1},

    1.65 X 10^{-3}\: dm^{-6}atm^{-2}\: mol^{2}

  • Option 3)

    \\1,4.1\times 10^{-2}dm^{-3}atm^{-1} \: mol,\\\\ 606dm^{6}atm^{2}mol^{-2}

  • Option 4)

    \\24.62dm^{3}atm\: mol^{-1},\\606.0 dm^{6}atm^{2}mol^{-2},\\1.65\times 10^{-3}dm^{-6}atm^{-2}mol^{2}

Answers (1)
A admin

 

Equilibrium in Chemical processes -

When the rates of forward and reverse reactions become equal the concentration of the reactants and the products remains constant. This is the stage of chemical equilibrium. 

-

 

 

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)

As we have learnt in chemical equilibriium

N_{2}(g)+O_{2}(g)\rightleftharpoons 2NO(g)

\frac{K_{p}}{K_{c}} = (RT)^{\Delta N}=\left ( RT \right )^{\circ}=1

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

\frac{K_{P}}{K_{C}}=(RT)^{'}= 24.62

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

\frac{K_{p}}{K_{c}} = \left ( RT \right )^{-2} = \frac{1}{\left ( RT \right )^{2}} = 1.65 \times 10^{-3}

 


Option 1)

1,24.62 \: dm^{3}atm\: mol^{-1},

606.0\: dm^{6}atm^{2}\: mol^{-2}

Option 2)

1,24.62\: dm^{3}atm\: mol^{-1},

1.65 X 10^{-3}\: dm^{-6}atm^{-2}\: mol^{2}

Option 3)

\\1,4.1\times 10^{-2}dm^{-3}atm^{-1} \: mol,\\\\ 606dm^{6}atm^{2}mol^{-2}

Option 4)

\\24.62dm^{3}atm\: mol^{-1},\\606.0 dm^{6}atm^{2}mol^{-2},\\1.65\times 10^{-3}dm^{-6}atm^{-2}mol^{2}

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