Chemistry, asked by abhilatasingh410, 2 months ago

JI
4. N.
2(g)
+O
2(g)
2NO
Nitrogen and oxygen react as illustrated by the equation above to form 1400 ml of nitric oxide gas. The volume
of oxygen which takes part in chemical reactions is :
(a) 800 ml
(b) 700 ml
(c) 600 ml
(d) 850 ml​

Answers

Answered by samamrita2009
1

Answer:

Explanation:

Chemical equilibrium occurs when a reaction and

its reverse reaction proceed at the same rate.

2

Concept of Equilibrium

• As a system approaches

equilibrium, both the

forward and reverse

reactions are occurring.

• At equilibrium, the

forward and reverse

reactions are proceeding

at the same rate. • Once equilibrium is

achieved, the amount of

each reactant and

product remains

constant.

The same equilibrium is reached whether we

start with only reactants (N2

and H2

) or with

only product (NH3 ).

Equilibrium is reached from either direction.  3

The Equilibrium Constant

• Consider the generalized reaction

The equilibrium expression for this

reaction would be

Kc =

[C]c

[D]d

[A]a

[B]b

aA + bB cC + dD

Since pressure is proportional to

concentration for gases in a closed

system, the equilibrium expression can

also be written

Kp

=

(PC

)c

(PD

)d

(PA

)a

(PB

)b

Chemical equilibrium occurs when

opposing reactions are proceeding at equal rates.

Since, in a system at equilibrium, both the forward

and reverse reactions are being carried out, we

write its equation with a double arrow.

Forward reaction:

N2O4 (g)  2 NO2 (g)

Rate Law:

Rate = kf [N2O4

]

N2O4 (g) 2 NO2 (g)

Reverse reaction:

2 NO2 (g)  N2O4 (g)

Rate Law:

Rate = kr [NO2

]2

4

Equilibrium Constant

• Therefore, at equilibrium

Ratef = Rater

kf [N2O4

] = kr [NO2

]2

• Rewriting this, it becomes

kf

kr

[NO2

]2

[N2O4

] =

Keq

=

kf

kr

[NO2

]2

[N2O4

]

= = a constant

Example

N2

(g) + 3 H2

(g) ⇌ 2 NH3

(g)

Write the equilibrium constant expression of the

following reaction:  

5

Relationship Between Kc and Kp

Plugging this into the expression for Kp

for each substance, the relationship

between Kc and Kp

becomes

where

Kp = Kc

(RT)n

n = (moles of gaseous product) - (moles of gaseous reactant)

From the Ideal Gas Law we know that:

PV = nRT and P = (n/V)RT = [A]RT

What Does the Value of K Mean?

• If K>>1, the reaction is

product-favored;

product predominates

at equilibrium.

• If K<<1, the reaction is

reactant-favored;

reactant predominates

at equilibrium.

*When 10-3 < K < 103, the reaction is considered to contain a

significant amount of both reactants and products at equilibrium.  

6

Direction of Chemical Equation and K

The equilibrium constant of a reaction in the

reverse direction is the reciprocal of the

equilibrium constant of the forward reaction.

Kc = = 0.212 at 100 C

[NO2

]2

[N2O4

] N2O4

(g) 2 NO2

(g)

Kc = = 4.72 at 100 C

[N2O4

]

[NO2

] N2O4 2

2 NO (g) 2

(g)

Kc = = 0.212 at 100 C

[NO2

]2

[N2O4

]

Kc = = (0.212)2

at 100 C

[NO2

]4

[N2O4

]2

4 NO2

2 N (g) 2O4

(g)

N2O4

(g) 2 NO2 ⇌ (g)

The equilibrium constant of a reaction that has

been multiplied by a number, is the equilibrium

constant raised to a power that is equal to that

number.

Stoichiometry and K

7

2 NOBr ⇌ 2 NO + Br2 K1 = 0.014

Br2 + Cl2 ⇌ 2 BrCl K2 = 7.2

2 NOBr + Cl2 ⇌ 2 NO + 2 BrCl

K3 = K1 × K2 = 0.014 × 7.2 = 0.10

The equilibrium constant for a net reaction

made up of two or more steps is the

product of the equilibrium constants for the

individual steps.

Multiple equilibria and K

Example

Consider the following reactions at 1200 K.

CO(g) + 3H2

(g) ⇌ CH4

(

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