The ratio of \({{K_{P}}\over{K_{C}}}\) for the reaction

$\mathrm{CO}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{CO}}_2\left(\mathrm{g}\right)$ $\mathrm{is:}$

\(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5}{[\mathrm{NO}]^4\left[\mathrm{H}_2 \mathrm{O}]^6\right.}\)
The balanced chemical equation corresponding to the above-mentioned expression is:

The equilibrium reaction that doesn't have equal values for K_c and K_p is: 

1. \(2NO(g) \rightleftharpoons N_2(g) + O_2(g)\)
2. \(SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)\)
3. \(H_2(g) + I_2(g) \rightleftharpoons 2HI (g)\)
4. \(2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)\)

For the reaction ${\mathrm{N}}_2\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_2\left(\mathrm{g}\right)\leftrightharpoons 2\mathrm{NO}\left(\mathrm{g}\right)$ the equilibrium constant is K₁. The equilibrium constant is K₂ for the reaction  $2\mathrm{NO}\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_2\left(\mathrm{g}\right)$ $\leftrightharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right)$ $$
The value of K for the reaction given below will be:
${\mathrm{NO}}_2\left(\mathrm{g}\right)\leftrightharpoons \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)$ $+{\mathrm{O}}_2\left(\mathrm{g}\right)$ $$

1.  $\frac{1}{4}$ $\left(4\right)$ ${\mathrm{K}}_1$ ${\mathrm{K}}_2$

2.  ${\left[\frac{1}{{\mathrm{K}}_1{\mathrm{K}}_2}\right]}^{1/2}$

3.  $\frac{1}{\left({\mathrm{K}}_1{\mathrm{K}}_2\right)}$

4.  $\frac{1}{\left(2{\mathrm{K}}_1{\mathrm{K}}_2\right)}$

The value of the equilibrium constant of the reaction 
 $\mathrm{HI}\left(\mathrm{g}\right)$ $\leftrightharpoons$ $\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)$ $+$ $\frac{1}{2}{\mathrm{I}}_2$ is 8.0.
The equilibrium constant of the reaction
 ${\mathrm{H}}_2\left(\mathrm{g}\right)$ $+$ ${\mathrm{I}}_2\left(\mathrm{g}\right)$ $\leftrightharpoons$ $2\mathrm{HI}\left(\mathrm{g}\right)\hspace{0.17em}$ will be-

For the following equilibrium, K_c= 6.3 × 10¹⁴ at 1000 K

$\mathrm{NO}\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_3$ $\left(\mathrm{g}\right)$ $\to$ ${\mathrm{NO}}_2\left(\mathrm{g}\right)$ $+$ ${\mathrm{O}}_2$ $\left(\mathrm{g}\right)$ $$

The value of K_c for the reverse reaction is:

$1.$ $2.33$ $\times$ ${10}^{-16}$
$2.$ $1.59$ $\times$ ${10}^{-15}$
$3.$ $2.67$ $\times$ ${10}^{-13}$
$4.$ $4.47$ $\times$ ${10}^{14}$

The value of $∆\mathrm{n}$ for the following reaction will be :

${\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{NH}}_3\left(\mathrm{g}\right)+\mathrm{HCl}\left(\mathrm{g}\right)$

For the reaction 2NOCl_(g)⇔2NO_(g)+Cl_2(g), K_C at 427$°$C is \(3\times 10^{-6} \ mol\ L^{-1}\). The value of K_p will be :

1. $1.72\times {10}^{-4}$

2. $7.50\times {10}^5$

3. $2.50\times {10}^{-5}$

4. $2.50\times {10}^{-4}$

PCl₅, PCl₃, and Cl₂ are at equilibrium at 500 K in a closed container and their concentrations are 0.8×10^–3 mol L^–1 , 1.2×10^–3 mol L^–1 and 1.2×10^–3 mol L^–1, respectively.
The value of  K_c  for the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) will be:
1. 1.8 × 10³ mol L^–1
2. 1.8 × 10³
3. 1.8 × 10^–3 mol L^–1
4. 0.55 × 10⁴

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