Question

Find out the value of $K_c$ for the given equilibria (i) and (ii) respectively:
$\left(i\right)2NOCl\left(g\right)\rightleftharpoons 2NO\left(g\right)+C{l}_2\left(g\right);K_P=1.8\times {10}^{-2}at500K$
$\left(ii\right)CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right);K_p=167at1073K$

($R=0.0831barLmo{l}^{-1}{K}^{-1}$)

• A. $4.38\times {10}^{-4}and1.87$
• B. $8.8\times {10}^{-4}and3.8$
• C. $4.38\times {10}^{4}and1.87$
• D. $8.8\times {10}^{4}and3.8$

Answer 1

The correct option is A $4.38\times {10}^{-4}and1.87$

$\left(i\right)2NOCl\left(g\right)\rightleftharpoons 2NO\left(g\right)+C{l}_2\left(g\right);K_P=1.8\times {10}^{-2}at500K$
$K_P=K_C(RT{)}^{△n_g}$
$\because △n_g=3-2=1$
$\therefore K_C=\frac{K_P}{RT}=\frac{1.8\times {10}^{-2}}{0.0831\times 500}=4.38\times {10}^{-4}$

$\left(ii\right)CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right);K_p=167at1073K$
$\because △n_g=1-0=1$
$K_C=\frac{K_P}{RT}=\frac{167}{0.0831\times 1073}=1.87$



Theory:

Relation between $K_p$ and $K_c$ :
$aA\left(g\right)+bB\left(g\right)\rightleftharpoons cC\left(g\right)+dD\left(g\right)$
For an ideal gas :
$PV=nRT$
$P=\left(\frac{n}{V}\right)RT$
Where $\frac{n}{V}$=C (molar concentration)
$C=\frac{P}{RT}$
$K_c=\frac{\left[C{]}_{eq}^c\right[D{]}_{eq}^d}{\left[A{]}_{eq}^a\right[B{]}_{eq}^b}$

$K_c=\frac{\frac{\left(P_C{)}_{eq}^c\right(P_D{)}_{eq}^d}{\left(RT{)}^c\right(RT{)}^d}}{\frac{\left(P_A{)}_{eq}^a\right(P_B{)}_{eq}^b}{\left(RT{)}^a\right(RT{)}^b}}$

$K_c=\frac{[P_C{]}_{eq}^c\times [P_D{]}_{eq}^d(RT{)}^{(a+b)-(c+d)}}{[P_A{]}_{eq}^a\times [P_B{]}_{eq}^b}$


By using the property $\frac{1}{a^x}\times \frac{1}{a^y}=\frac{1}{a^{x+y}}$

$K_C=K_P(RT{)}^{-\Delta n_g}\Rightarrow K_P=K_C(RT{)}^{\Delta n_g}$
$\Delta n_g$ = (number of moles of gaseous products) - (number of moles of gaseous reactants)
If $\Delta n_g=0$ then $K_c=K_p$