Question

5 . For the following equilibrium, K =6.3 x 1014 at 1000 K. NO(g)+O3 —–>NO2(g) + O2(g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?

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Answer 1

Answer:

The value of the equilibrium constant for the reverse reaction is obtained from the reciprocal of the value of the equilibrium constant of the forward reaction.

Hence, KC(reverse)=KC1.

Substitute values in the above expression.

KC(reverse)=6.3×10141=1.59×10-¹⁵

Answer 2

The chemical reaction between NO and ozone to give nitrogen dioxide and oxygen gas.

The chemical reaction is as follows.

NO(g)\quad +\quad { O }_{ 3 }(g)\quad \rightarrow \quad { NO }_{ 2 }(g)\quad +\quad { O }_{ 2 }(g)NO(g)+O

3

(g)→NO

2

(g)+O

2

(g)

It is forward reaction. The equilibrium constant of the forward reaction { K }_{ c }\quad =\quad 6.3\quad \times \quad 1{ 0 }^{ 14 }K

c

=6.3×10

14

For, Forward reaction =\quad \frac { 1 }{ { K }_{ c } }=

K

c

1

{ K }_{ c }\quad =\quad 6.3\quad \times \quad { 10 }^{ 14 }K

c

=6.3×10

14

Substitute the given values,

For backward reaction =\quad \frac { 1 }{ 6.3\quad \times \quad { 10 }^{ 14 } } \quad =\quad 1.59\quad \times \quad { 10 }^{ -15 }=

6.3×10

14

1

=1.59×10

−15

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