Reaction Quot...

Reaction Quotient

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Q. The free energy of formation of NO is

78 k J m o l^{- 1}

at the temperature of

1000 K

. What is the equilibrium constant for this reaction at

1000 K

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

$8.4 \times 10^{- 5}$
$7.1 \times 10^{- 9}$
$1.7 \times 10^{- 19}$
$4.2 \times 10^{- 10}$

View Solution

Q. For the following reaction:

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

0.4 m o l

P C l_{5} (g)

0.2 m o l

P C l_{3} (g)

and

0.6 m o l

C l_{2} (g)

are taken in a

1 L

flask.
If

K_{c} = 0.2

, then the direction in which reaction proceeds is :

Towards reactant side
Towards products side
Reaction is at equilibrium
None of the above

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N_{2} O_{4} (g) ⇌ 2 N O_{2} (g); K_{C} = 4

This reversible reaction is studied graphically as shown in figure. Select correct statements out of I, II and III
(I) Reaction quotient has maximum value at point A.
(II) Reaction proceeds from left to right at a point when,

[N_{2} O_{4}] = [N O_{2}] = 0.1 M

.
(III)

K_{c} = Q

, when point D or F is reached

I, II
II, III
I, III
I, II, III

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Q. Given four reactions and their equilibrium constants.

N_{2} O_{2} ⇌ 2 N O, K_{1}; \frac{1}{2} N_{2} + \frac{1}{2} O_{2} ⇌ N O, K_{2};

2 N O ⇌ N_{2} + O_{2}, K_{3}; N O ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}, K_{4}

Correct relation between

K_{1}, K_{2}, K_{3} a n d K_{4}

is:

$(K_{4})^{2} = K_{3}$
$\sqrt{K_{3}} \times K_{2} = 1$
both a and b
None of the above

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Q. The Reaction Quotient (Q) for the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

is given by

Q = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}

, the reaction will shift in forward direction if :

$Q = K_{c}$
$Q < K_{c}$
$Q > K_{c}$
$Q = 0$

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3g of $H_{2}$ and 24g of $O_{2}$ are present in a gaseous mixture at constant temperature and pressure. Find the partial pressure of hydrogen.

$\frac{1}{2}$ of total pressure
$\frac{2}{3}$ of total pressure
$\frac{1}{3}$ of total pressure
$\frac{3}{2}$ of total pressure

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Q. Consider the following reaction and its equilibrium constant:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) K c = 5.85 \times 10^{- 3}

A reaction mixture at a particular time, t contains:

[N O_{2}] = 0.025 M a n d [N_{2} O_{4}] = 0.033 M .

Calculate

Q c

, and determine the direction in which the reaction will proceed.

$1.9 \times 10^{- 3}$ , the reaction will proceed in forward direction
$1.9 \times 10^{- 2}$ , the reaction will proceed in backward direction
$5.85 \times 10^{- 3}$ , it will be equal to the equilibrium constant
None of the above

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Q. At a certain temperature (T), the equilibrium constant

(K_{c})

is 1 for the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

If 2 moles of

N_{2},

4 moles of

H_{2},

6 moles of

N H_{3}

and 3 moles of inert gas are introduced into a two litre rigid vessel at constant temperature T. It has been found that equilibrium concentration of

H_{2}

and

N H_{3}

are equal then what is the equilibrium concentration of

N_{2}

(in M)?

View Solution

Q. The free energy of formation of NO is

78 k J m o l^{- 1}

at the temperature of

1000 K

. What is the equilibrium constant for this reaction at

1000 K

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

$8.4 \times 10^{- 5}$
$7.1 \times 10^{- 9}$
$4.2 \times 10^{- 10}$
$1.7 \times 10^{- 19}$

View Solution

Q. Two equilibria exist simultaneously in a vessel at

25^{\circ} C

\begin{matrix} N O (g) & + & N O_{2} (g) & K_{P 1} = 0.4 a t m^{- 1} ⇌ & N_{2} O_{3} (g) \end{matrix}

\begin{matrix} 2 N O_{2} (g) & K_{P 2} = 8 a t m^{- 1} ⇌ & N_{2} O_{4} (g) \end{matrix}

If initially only

N O

and

N O_{2}

are present in a 3:5 mole ratio and the final pressure at equilibrium of

N O_{2}

at 0.5 atm, then calculate total pressure (in atm) at equilibrium.

View Solution

Q. Calculate the value of

Q_{c}

value, for the given reaction below. Also, given the concentration of all the species involved at a certain time:

C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g) [C O_{2}] = 2 M [H_{2}] = 2 M [C O] = 1 M [H_{2} O] = 1 M

K_{c} = 1

, in which direction the reaction will shift to attain equilibrium?

$Q_{C} = 2$ and forward
$Q_{C} = 2$ and backward
$Q_{C} = 4$ and forward
$Q_{C} = 4$ and backward

View Solution

Q. The equilibrium constant

K_{P}

for the reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2}

(g) is

4.5

.
What would be the average molar mass (in g⁄mol) of an equilibrium mixture of

N_{2} O_{4}

and

N O_{2}

formed by the dissociation of pure

N_{2} O_{4}

at a total pressure of 2 atm?

$69$
$57.5$
$80.5$
$85.5$

View Solution

Q. The standard Gibbs free energy change for the reaction:

2 N O B r (g) ⇌ 2 N O (g) + B r_{2} (g)

+ 12.50 kJ/mole

27^{\circ} C

. What is the equilibrium constant for the reaction at

27^{\circ} C

$10^{5}$
$10^{- 5}$
$e^{5}$
$e^{- 5}$

View Solution

Q. Consider the following reaction and its equilibrium constant:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) K c = 5.85 \times 10^{- 3}

A reaction mixture at a particular time, t contains:

[N O_{2}] = 0.025 M a n d [N_{2} O_{4}] = 0.033 M .

Calculate

Q c

, and determine the direction in which the reaction will proceed.

$1.9 \times 10^{- 3}$ , the reaction will proceed in forward direction
$1.9 \times 10^{- 2}$ , the reaction will proceed in backward direction
$5.85 \times 10^{- 3}$ , it will be equal to the equilibrium constant
None of the above

View Solution

Q. The standard Gibbs free energy change for the reaction:

2 N O B r (g) ⇌ 2 N O (g) + B r_{2} (g)

+ 12.50 kJ/mole

27^{\circ} C

. What is the equilibrium constant for the reaction at

27^{\circ} C

$10^{5}$
$10^{- 5}$
$e^{5}$
$e^{- 5}$

View Solution

Q. At a certain temperature, equilibrium constant

K_{c} = 4 \times 10^{- 2}

for the reaction

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

If we take

1.5 m o l

of NO and

2 m o l

each of

N_{2} a n d O_{2}

5 L

vessel, what would be the equilibrium concentration of

N O

(in

m o l / L

View Solution

Q. The free energy of formation of NO is

78 k J m o l^{- 1}

at the temperature of

1000 K

. What is the equilibrium constant for this reaction at

1000 K

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

$8.4 \times 10^{- 5}$
$7.1 \times 10^{- 9}$
$4.2 \times 10^{- 10}$
$1.7 \times 10^{- 19}$

View Solution