For the equilibrium reaction:

2NOCl(g) ⇌ 2NO(g) + Cl₂(g)

The equilibrium constant in terms of concentration (Kc) is 3.0 × 10⁻⁶ at 1000 K.
Calculate the value of Kp for the reaction at this temperature.

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. \(2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)\)

Find the equilibrium constant K for the reaction:

NO₂(g) ⇌ ½N₂(g) + O₂(g)

Given:

N₂(g) + O₂(g) ⇌ 2NO(g), K₁

2NO(g) + O₂(g) ⇌ 2NO₂(g), K₂

Find the equilibrium constant for the reaction:

H₂(g) + I₂(g) ⇌ 2HI(g)

Given that:

HI(g) ⇌ ½H₂(g) + ½I₂(g)  K = 8

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. $2.50\times {10}^{-4}$