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:

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

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

One mole of H₂O and one mole of CO are taken in a 10 L vessel and heated to
725 K. At equilibrium, 40% of water(by mass) reacts with CO according to the equation,

H₂O _(g) + CO _(g)  $\leftrightharpoons$H_2 (g) + CO_2 (g)

The equilibrium constant for the above-mentioned reaction would be:

The equilibrium pressure of C₂H₆ when it is placed in a flask at 4.0 atm pressure at  899 K would be:

C₂H_6 (g)  $\leftrightharpoons$ C₂H_4 (g) + H_2 (g)

( K_p = 0.04 atm at 899 K)

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K.
After equilibrium was attained, a concentration of PCl₅ 
was found to be 0.5 × 10^–1 mol L^–1. If the value of
 K_c is 8.3 × 10^–3 mol L^–1, the concentrations of
PCl₃ and Cl₂ at equilibrium would be:

PCl_5 (g)  $\leftrightharpoons$PCl_3 (g) + Cl_2(g)
 

Match the following equilibria with the corresponding condition.

Codes

​Some reactions are written below in Column I and their equilibrium constants
in terms of K_c are written in Column II.
Match the following reactions with the corresponding equilibrium constant.

Codes

Match the standard free energy of the reaction with the corresponding equilibrium constant.

Codes

Match the following species with the corresponding conjugate acid.

Codes

Match the graphical variations with their descriptions given above and identify the correct codes below: