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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)

best_answer

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 K_p , 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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