NEET Chemistry: Equilibrium PYQs Chapterwise Practice

Q 1 :
For the reaction $A_{(g)} ⇌ 2 B_{(g)}$ , the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.

[Given: $R = 0.0831$ $L$ $atm$ ${mol}^{- 1}$ $K^{- 1}$ ]

$K_{p}$ for the reaction at 1000 K is: [2025]

0.033
0.021
83.1
$2.077 \times 10^{5}$

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(1)

For reaction, $A_{(g)} ⇌ 2 B_{(g)}$

R = 0.0831 L atm ${mol}^{- 1}$ $K^{- 1}$ , T = 1000 K

$K_{c} = \frac{k_{f}}{k_{b}} = \frac{k_{f}}{2500 k_{f}} = \frac{1}{2500}$

As $K_{p} = K_{c} {(R T)}^{Δ n_{g}}$

$K_{p} = \frac{1}{2500} {(0.0831 \times 1000)}^{1} = 0.033$ $[As, Δ n_{g} = 2 - 1 = 1]$

Q 2 :
At a given temperature and pressure the equilibrium constant values for the equilibria are given below.

$3 A_{2} + B_{2} ⇌ 2 A_{3} B, K_{1}$

$A_{3} B ⇌ \frac{3}{2} A_{2} + \frac{1}{2} B_{2}, K_{2}$

The relation $K_{1}$ and $K_{2}$ is [2024]

$K_{1}^{2} = 2 K_{2}$
$K_{2} = K_{1} / 2$
$K_{1} = \frac{1}{\sqrt{K_{2}}}$
$K_{2} = \frac{1}{\sqrt{K_{1}}}$

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(4)

$K_{1} = \frac{{[A_{3} B]}^{2}}{{[A_{2}]}^{3} [B_{2}]}$ ...(i)

$K_{2} = \frac{{[A_{2}]}^{3 / 2} {[B_{2}]}^{1 / 2}}{[A_{3} B]}$ ...(ii)

Taking square root on both sides in eq. (i), we get

$\sqrt{K_{1}} = \frac{[A_{3} B]}{{[A_{2}]}^{3 / 2} {[B_{2}]}^{1 / 2}}$

$\Rightarrow [A_{3} B] = \sqrt{K_{1}} {[A_{2}]}^{3 / 2} {[B_{2}]}^{1 / 2}$

Putting the value of $[A_{3} B]$ in eq. (ii), we get

$K_{2} = \frac{1}{\sqrt{K_{1}}}$

Q 3 :
$3 O_{2 (g)} ⇌ 2 O_{3 (g)}$

For the above reaction at 298 K, $K_{c}$ is found to be $3.0 \times 10^{- 59}$ . If the concentration of $O_{2}$ at equilibrium is 0.040 M, then the concentration of $O_{3}$ in M is [2022]

$4.38 \times 10^{- 32}$
$1.9 \times 10^{- 63}$
$2.4 \times 10^{31}$
$1.2 \times 10^{21}$

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(1)

$\begin{matrix} 3 O_{2 (g)} & ⇌ & 2 O_{3 (g)} \\ Initial conc . & 1 & 0 \\ At equilibrium & 1 - 3 x & 2 x \end{matrix}$

$K_{c} = \frac{{[O_{3}]}^{2}}{{[O_{2}]}^{3}} = 3 \times 10^{- 59}$

Given: $[O_{2}] = 0.040$ M

$K_{c} = \frac{{[O_{3}]}^{2}}{{(0.040)}^{3}} = 3 \times 10^{- 59}$

${[O_{3}]}^{2} = 1.92 \times 10^{- 63}$

$[O_{3}] = 4.38 \times 10^{- 32}$ M

Q 4 :
The equilibrium constants of the following are

$N_{2} + 3 H_{2}$ $⇌ 2 {NH}_{3}; K_{1}$

$N_{2} + O_{2} ⇌ 2 NO; K_{2}$

$H_{2} + \frac{1}{2} O_{2} ⇌ H_{2} O; K_{3}$

The equilibrium constant (K) of the reaction:

$2 {NH}_{3} + \frac{5}{2} O_{2} \overset{k}{⇌} 2 NO + 3 H_{2} O$ will be [2017, 2007, 2003]

$\frac{K_{2} K_{3}^{3}}{K_{1}}$
$\frac{K_{2} K_{3}}{K_{1}}$
$\frac{K_{2}^{3} K_{3}}{K_{1}}$
$\frac{K_{1} K_{3}^{3}}{K_{2}}$

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(1)

From the given equations,

$2 {NH}_{3} ⇌ N_{2} + 3 H_{2}; \frac{1}{K_{1}}$ ...(i)

$N_{2} + O_{2} ⇌ 2 NO; K_{2}$ ...(ii)

$3 H_{2} + \frac{3}{2} O_{2} ⇌ 3 H_{2} O; K_{3}^{3}$ ...(iii)

By adding equations (i), (ii) and (iii), we get

$2 {NH}_{3} + \frac{5}{2} O_{2} \overset{k}{⇌} 2 NO + 3 H_{2} O, K = \frac{K_{2} K_{3}^{3}}{K_{1}}$

Q 5 :
If the equilibrium constant for

$N_{2 (g)} + O_{2 (g)} ⇌ 2 {NO}_{(g)}$ is K, the equilibrium constant for $\frac{1}{2} N_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ {NO}_{(g)}$ will be [2015]

$\frac{1}{2} K$
$K$
$K^{2}$
$K^{1 / 2}$

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(4)

If the reaction is multiplied by $\frac{1}{2}$ , then new equilibrium constant, $K' = K^{1 / 2}$ .

Topic Question Set

Q 1 :
For the reaction $A_{(g)} ⇌ 2 B_{(g)}$ , the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.

[Given: $R = 0.0831$ $L$ $atm$ ${mol}^{- 1}$ $K^{- 1}$ ]

$K_{p}$ for the reaction at 1000 K is: [2025]

Q 2 :
At a given temperature and pressure the equilibrium constant values for the equilibria are given below.

$3 A_{2} + B_{2} ⇌ 2 A_{3} B, K_{1}$

$A_{3} B ⇌ \frac{3}{2} A_{2} + \frac{1}{2} B_{2}, K_{2}$

The relation $K_{1}$ and $K_{2}$ is [2024]

Q 3 :
$3 O_{2 (g)} ⇌ 2 O_{3 (g)}$

For the above reaction at 298 K, $K_{c}$ is found to be $3.0 \times 10^{- 59}$ . If the concentration of $O_{2}$ at equilibrium is 0.040 M, then the concentration of $O_{3}$ in M is [2022]

Q 5 :
If the equilibrium constant for

$N_{2 (g)} + O_{2 (g)} ⇌ 2 {NO}_{(g)}$ is K, the equilibrium constant for $\frac{1}{2} N_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ {NO}_{(g)}$ will be [2015]

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Topic Question Set

Q 1 : For the reaction A(g)⇌2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given: R=0.0831 L atm mol-1 K-1] Kp for the reaction at 1000 K is: [2025]

Q 2 : At a given temperature and pressure the equilibrium constant values for the equilibria are given below. 3A2+B2⇌2A3B,K1 A3B⇌32A2+12B2,K2 The relation K1 and K2 is [2024]

Q 3 : 3O2(g)⇌2O3(g) For the above reaction at 298 K, Kc is found to be 3.0×10-59. If the concentration of O2 at equilibrium is 0.040 M, then the concentration of O3 in M is [2022]

Q 4 : The equilibrium constants of the following are N2+3H2⇌2NH3;K1 N2+O2⇌2NO;K2 H2+12O2⇌H2O;K3 The equilibrium constant (K) of the reaction: 2NH3+52O2⇌k2NO+3H2O will be [2017, 2007, 2003]

Q 5 : If the equilibrium constant for N2(g)+O2(g)⇌2NO(g) is K, the equilibrium constant for 12N2(g)+12O2(g)⇌NO(g) will be [2015]

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Q 1 :
For the reaction $A_{(g)} ⇌ 2 B_{(g)}$ , the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.

[Given: $R = 0.0831$ $L$ $atm$ ${mol}^{- 1}$ $K^{- 1}$ ]

$K_{p}$ for the reaction at 1000 K is: [2025]

Q 2 :
At a given temperature and pressure the equilibrium constant values for the equilibria are given below.

$3 A_{2} + B_{2} ⇌ 2 A_{3} B, K_{1}$

$A_{3} B ⇌ \frac{3}{2} A_{2} + \frac{1}{2} B_{2}, K_{2}$

The relation $K_{1}$ and $K_{2}$ is [2024]

Q 3 :
$3 O_{2 (g)} ⇌ 2 O_{3 (g)}$

For the above reaction at 298 K, $K_{c}$ is found to be $3.0 \times 10^{- 59}$ . If the concentration of $O_{2}$ at equilibrium is 0.040 M, then the concentration of $O_{3}$ in M is [2022]

Q 5 :
If the equilibrium constant for

$N_{2 (g)} + O_{2 (g)} ⇌ 2 {NO}_{(g)}$ is K, the equilibrium constant for $\frac{1}{2} N_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ {NO}_{(g)}$ will be [2015]