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. $2.50\times {10}^{-4}$

For the following reaction, 
 H$$₂(g)+I₂(g)⇌2HI(g) $at$ $250°C,$

The effect on the state of equilibrium on doubling the volume of the system will be:

In the reaction A(g) + 2B(g) ⇌ 2C(g) + D(g), the initial concentration of B is twice that of A and, at equilibrium, the concentrations of A and D are equal. The value of the equilibrium constant will be:

The number of ${\mathrm{H}}^{\oplus }$ ions present in 1 ml of a solution whose pH= 4 , is given as:

($N_A=$ $Avogadro\text{'}s$ $number$)

1. ${10}^{-7}{N}_A$

2. ${10}^{-8}{N}_A$

3. ${10}^{-16}{N}_A$

4. ${10}^{-14}{N}_A$

If the pH of an acidic buffer is 5.7 and ${\mathrm{pK}}_{\mathrm{a}}$ is 5 then the ratio of $\frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$ will be:

Assertion: At equilibrium, ${\mathrm{K}}_{}={\mathrm{Q}}_{\mathrm{c}}.$

Reason: When ${\mathrm{K}}_{\mathrm{c }}> {\mathrm{Q}}_{\mathrm{c}},$ the reaction mainly goes in the forward direction.

Assertion: $H^{-}$ acts as both Bronsted Lowry base as well as Lewis base.

Reason: It acts as both $H^{\oplus }$ acceptor and a lone pair donor.

1. Both assertion & reason are true and the reason is the correct explanation of the assertion.

2. Both assertion & reason are true but the reason is not the correct explanation of the assertion.

3. Assertion is a true statement but the reason is false.

4. Both assertion and reason are false statements.

The concentration of $\left[{\mathrm{H}}^{+}\right]$ ion in a solution containing 0.1 M HCN and 0.2 M NaCN will be:

(${\mathrm{K}}_{\mathrm{a}}$ for HCN = $6.2\times {10}^{-10}$)

1. 3.1$\times {10}^{10}$

2. $6.2\times {10}^5$

3.$$ $6.2\times {10}^{-10}$

4. $3.1\times {10}^{-10}$

Given a hypothetical reaction :
${\mathrm{AB}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{B}}_2\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{AB}}_3\left(\mathrm{g}\right);$ $∆\mathrm{H}=-\mathrm{x}$ $\mathrm{kJ}$
More AB₃​ could be produced at equilibrium by :

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