If the value of an equilibrium constant for a particular reaction is 1.6 × 10 12 , then at equilibrium the system will contain:

Similar amounts of reactants and products

For the reaction: 2 NO 2 g ⇌ 2 NO g + O 2 g , K c = 1 .8 × 10 - 6 at 184 o C , R = 0 .0831 J / molK When K p and K c are compared at 184 o C. It is found that

Whether K p is greater than, less than or equal to K c depends upon the total gas pressure.

EASY

Earn 100

0.6 mole of $P C l_{5}, 0.3$ mole of $P C l_{3}$ and 0.4 mole of $C l_{2}$ are taken in a 1 L flask to obtain the following equilibrium
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
If the equilibrium constant $K_{c}$ for the reaction is 0.15. Predict the direction of the reaction.

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Important Questions on Equilibrium

EASY

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

Consider the following reversible chemical reactions:

A_{2} (g) + B_{2} (g) \overset{k_{1}}{⇌} 2 A B (g)

.....(1)

6 A B (g) \overset{k_{2}}{⇌} 3 A_{2} (g) + 3 B_{2} (g)

.....(2)

The relation between

K_{1}

and

K_{2}

is:

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

$4.5$ moles each of hydrogen and iodine is heated in a sealed ten litre vessel.At equilibrium, $3$ moles of $HI$ were found. The equilibrium constant for $H_{2} (g) + I_{2} (g) ⇌ 2 {HI}_{(g)}$ is ........

HARD

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

What are the values of

\frac{K_{p}}{K_{c}}

for the following reactions at

300 K

respectively?

(At

300 K, RT = 24 .62 {dm}^{3} atm {mol}^{- 1}

)

N_{2} (g) + O_{2} (g) ⇌ 2 NO (g); \frac{K_{p}}{K_{c}} = x_{1}

N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g); \frac{K_{p}}{K_{c}} = x_{2}

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g); \frac{K_{p}}{K_{c}} = x_{3}

HARD

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

Consider the equilibrium

X_{2} + Y_{2} ⇌ P

. Find the stoichiometric coefficient of the

P

using the data given in the following table:

$X_{2} / m o l L^{- 1}$	$Y_{2} / m o l L^{- 1}$	$P / m o l L^{- 1}$
$1.14 \times 10^{- 2}$	$0.12 \times 10^{- 2}$	$2.52 \times 10^{- 2}$
$0.92 \times 10^{- 2}$	$0.22 \times 10^{- 2}$	$3.08 \times 10^{- 2}$

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

A sample of

HI (g)

is placed in a flask at a pressure of

0.2 atm .

At equilibrium, partial pressure of

HI (g)

0.04 atm .

What is

Kp

for the given equilibrium ?

2 HI (g) ⇌ H_{2} (g) + I_{2} (g)

EASY

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

If the value of an equilibrium constant for a particular reaction is

1.6 \times 10^{12}

, then at equilibrium the system will contain:

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

In a chemical reaction,

A + 2 B \overset{K}{⇌} 2 C + D

, the initial concentration of

B

was

1.5

times of the concentration of

A

, but the equilibrium concentrations of

A

and

B

were found to be equal. The equilibrium constant

(K)

for the chemical reaction is:

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

The equilibrium constant

(K_{c})

of two reactions

H_{2} + I_{2} ⇌ 2 HI and N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}

are

50 and 1000

, respectively. The equilibrium constant of the reaction

N_{2} + 6 HI ⇌ 2 {NH}_{3} + 3 I_{2}

is closest to:

EASY

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

Burning the coal is represented as

C_{(s)} + O_{2 (g)} ⟶ {CO}_{2 (g)} .

The rate of this reaction is increased by

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

For the reaction:

2 {NO}_{2 (g)} ⇌ 2 {NO}_{(g)} + O_{2 (g)},

(K_{c} = 1 .8 \times 10^{- 6} at 184^{o} C), (R = 0 .0831 J / molK)

When

K_{p}

and

K_{c}

are compared at

184^{o} C.

It is found that

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

The equilibrium constants of the following are:

\begin{matrix} 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} \to H_{2} O & K_{3} \end{matrix}

The equilibrium constant (

K

) of the reaction:

2 N H_{3} + \frac{5}{2} O_{2} \overset{K}{⇌} 2 N O + 3 H_{2} O

, will be:

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

The value of $K_{c} us 64 at 800 K$ for the reaction

$N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)$

The value of $K_{c}$ for the following reaction is :

${NH}_{3} (g) \to \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

The equilibrium constant

K_{c}

of the reaction,

2 A ⇌ B + C is 0 .5 at 25^{o}

and

1 atm

. The reaction will proceed in the backward direction when concentrations

[A], [B] and [C]

are, respectively -

EASY

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

If the equilibrium constant for

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

is K, the equilibrium constant for

\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ N O (g)

will be:

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

Consider the equilibria

(i) a n d (i i)

with equilibrium constants

K_{1} a n d K_{2}

, respectively

{SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {SO}_{3} (g)

.... (i)

2 {SO}_{3} (g) ⇌ 2 {SO}_{2} (g) + O_{2} (g)

.... (ii)

K_{1} a n d K_{2}

are related as

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

At a certain temperature, only 50% HI is dissociated into H₂ and I₂ at equilibrium. The equilibrium constant is :

EASY

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

For the reaction

{SO}_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ {SO}_{3} (g)

, if

K_{P} = K_{C} {(RT)}^{x}

where the symbols have usual meaning then the value of

x

is:
(assuming ideality)

EASY

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

For the reaction:

${SO}_{2 (g)} + \frac{1}{2} O_{2 (g)} \overset{}{⇌} {SO}_{3 (g)}$

$\frac{K_{P}}{K_{C}}$ is

HARD

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

The reaction

${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$

is in equilibrium in a closed vessel at $298 K .$ The partial pressure (in atm) of ${CO}_{2} (g)$ in the reaction vessel is closest to:

[Given: The change in Gibbs energies of formation at $298 K$ and $1$ bar for

$CaO (s) = - 603.501 kJ {mol}^{- 1}$

${CO}_{2} (g) = - 394.389 kJ {mol}^{- 1}$

${CaCO}_{3} (s) = - 1128.79 kJ {mol}^{- 1}$

Gas constant $R = 8.314 J K^{- 1} {mol}^{- 1}$ ]

EASY

Chemistry>Physical Chemistry>Equilibrium>Law of Chemical Equilibrium and Equilibrium Constants

The decay profiles of three radioactive species

A, B

and

C

are given below:

Question Image

These profiles imply that the decay constants

k_{A}, k_{B}

and

k_{C}

follow the order

0.6 mole of PCl5,0.3 mole of PCl3 and 0.4 mole of Cl2 are taken in a 1 L flask to obtain the following equilibrium PCl5(g)⇌PCl3(g)+Cl2(g) If the equilibrium constant Kc for the reaction is 0.15. Predict the direction of the reaction.

Important Questions on Equilibrium

Learn and Prepare for any exam you want !

0.6 mole of $P C l_{5}, 0.3$ mole of $P C l_{3}$ and 0.4 mole of $C l_{2}$ are taken in a 1 L flask to obtain the following equilibrium
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
If the equilibrium constant $K_{c}$ for the reaction is 0.15. Predict the direction of the reaction.