If the equilibrium constant for N₂(g) + O₂ (g) ⇄ 2NO(g) is K, the equilibrium constant for \(\frac{1}{2}\)N₂(g) + \(\frac{1}{2}\)O₂(g) ⇄ NO(g) will be?

1. $K^{\frac{1}{2}}$

2. $\frac{1}{2}K$

3. K

4. $K^2$

The value of the equilibrium constant for a particular reaction is 1.6 × 10¹². When the system is in equilibrium, it will include:
1. All reactants
2. Mostly reactants
3. Mostly products
4. Similar amounts of reactants and products

For a given exothermic reaction, K_p and K_p^’ are the equilibrium constants at temperatures T₁ and T₂ respectively. Assuming that the heat of reaction is constant in temperatures range between T₁ and T₂, it is a readily observation that:
1. $K_p>K_p^{\text{'}}$
2. $K_p<K_p^{\text{'}}$
3. $K_p=K_p^{\text{'}}$
4. $K_p=\frac{1}{K_p^{\text{'}}}$

Given that the equilibrium constant for the reaction 

$2S{O}_{2\left(g\right)}$ $+$ $O_{2\left(g\right)}$ $\leftrightharpoons$ $2S{O}_{3\left(g\right)}$ $$

has a value of 278 at a particular temperature, the value of the equilibrium constant for the following reaction at the same temperature will be:

$S{O}_{3\left(g\right)}$ $$ $\leftrightharpoons$ $S{O}_{2\left(g\right)}$ $+$ $1/2$ $O_{2\left(g\right)}$ $$

1.  $3.6$ $\times$ ${10}^{-3}$ $$

2.  $6.0$ $\times$ ${10}^{-2}$ $$

3.  $1.3$ $\times$ ${10}^{-5}$ $$

4.  $1.8$ $\times$ ${10}^{-3}$ $$

Given reaction:  A₂(g) + B₂(g)  ⇋ 2AB(g)  

At equilibrium, the concentrations of  A₂ = 3.0×10^–3 M;  B₂ = 4.2×10^–3 M and AB = 2.8×10^–3M. If the reaction takes place in a sealed vessel at  527 °C, then the value of  K_C will be: 

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

The reaction-

$2A+B\left(g\right)$ $\rightleftharpoons$ $3C\left(g\right)+D\left(g\right)$

begins with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression:

1. $\left[{(0.75)}^3\right(0.25\left)\right]÷\left[{(0.50)}^2\right(0.75\left)\right]$

2. $\left[{(0.75)}^3\right(0.25\left)\right]÷\left[{(0.50)}^2\right(0.25\left)\right]$

3. $\left[{(0.75)}^3\right(0.25\left)\right]÷\left[{(0.75)}^2\right(0.25\left)\right]$

4. $\left[{(0.75)}^3\right(0.25\left)\right]÷\left[{(1.00)}^2\right(1.00\left)\right]$

The dissociation constants for acetic acid and HCN at 25 °C are 1.5 x 10^-5 and 4.5 x 10^-10, respectively. The equilibrium constant for the equilibrium, 
CN^- + CH₃COOH  ⇌  HCN + CH₃COO^- 
would be:
1. $3.0\times {10}^5$
2. $3.0\times {10}^{-5}$
3. $3.0\times {10}^{-4}$
4. $3.0\times {10}^4$

The dissociation equilibrium of a gas AB₂ can be represented as 
 $2A{B}_2\left(g\right)$ $\leftrightharpoons$ $2AB\left(g\right)$ $+$ $B_2\left(g\right)$ $$
The degree of dissociation is ‘x’ and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K_P and total pressure p is:
1. (2KP/p)
2. (2Kp /p)^1/3
3. (2KP/p)^1/2
4. (KP/P)