Question

The following reactions occur at 500 K. Arrange them in order of increasing tendency to proceed to completion.
(1)- $2NOCl(g)\rightleftharpoons 2NO(g)+C{{l}_{2}}(g);\text{ }{{K}_{p}}=1.7\text{ x 1}{{\text{0}}^{-2}}$
(2)- ${{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g);\text{ }{{K}_{p}}=1.5\text{ x 1}{{\text{0}}^{-3}}$
(3)- $2S{{O}_{3}}(g)\rightleftharpoons S{{O}_{2}}(g)+{{o}_{2}}(g);\text{ }{{K}_{p}}=1.3\text{ x 1}{{\text{0}}^{-5}}$
(4)- $2N{{O}_{2}}(g)\rightleftharpoons 2NO(g)+{{O}_{2}}(g);\text{ }{{K}_{p}}=5.9\text{ x 1}{{\text{0}}^{-5}}$

(a)- 2 < 1 < 4 < 3
(b)- 3 < 4 < 2 < 1
(c)- 1 < 3 < 4 < 2
(d)- 4 < 3 < 1 < 2

Answer 1

Hint: We can predict the extent of reaction by three ways: If the ${{K}_{p}}\text{ or }{{K}_{c}}$ is large ( > ${{10}^{3}}$) then the reaction is forward, if the ${{K}_{p}}\text{ or }{{K}_{c}}$is low ( < ${{10}^{-3}}$) then the reaction is backward, if the ${{K}_{p}}\text{ or }{{K}_{c}}$is intermediate (${{10}^{3}}-{{10}^{-3}}$) then the products and reactants are comparable.


Complete step by step answer:
In equilibrium reaction ${{K}_{p}}\text{ or }{{K}_{c}}$known as equilibrium constant, can be used to predict extent of the reaction. Or we can say that the larger the value of the equilibrium constant greater the extent of the reaction to proceed to form products.
There are some generalizations used to predict the extent of reaction:
If the ${{K}_{p}}\text{ or }{{K}_{c}}$ is large ( > ${{10}^{3}}$) then the reaction is forward.
If the ${{K}_{p}}\text{ or }{{K}_{c}}$is low ( < ${{10}^{-3}}$) then the reaction is backward.
If the ${{K}_{p}}\text{ or }{{K}_{c}}$is intermediate (${{10}^{3}}-{{10}^{-3}}$) then the products and reactants are comparable.
So in the question, we are given four reactions whose equilibrium constants are as follows:
(1)- $2NOCl(g)\rightleftharpoons 2NO(g)+C{{l}_{2}}(g);\text{ }{{K}_{p}}=1.7\text{ x 1}{{\text{0}}^{-2}}$
(2)- ${{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g);\text{ }{{K}_{p}}=1.5\text{ x 1}{{\text{0}}^{-3}}$
(3)- $2S{{O}_{3}}(g)\rightleftharpoons S{{O}_{2}}(g)+{{o}_{2}}(g);\text{ }{{K}_{p}}=1.3\text{ x 1}{{\text{0}}^{-5}}$
(4)- $2N{{O}_{2}}(g)\rightleftharpoons 2NO(g)+{{O}_{2}}(g);\text{ }{{K}_{p}}=5.9\text{ x 1}{{\text{0}}^{-5}}$
And we know that greater the value of equilibrium constant greater the tendency for completion. From the above data, the reaction 1 has the highest ${{K}_{p}}$and the reaction 3 has the least ${{K}_{p}}$.
So, the order will be: 3 < 4 < 2 < 1

Therefore, the correct answer is an option (b)- 3 < 4 < 2 < 1


Note: There is another term used to predict the direction of the reaction, reaction quotient represented as Q. If Q is equal to K, then the reaction is in equilibrium if Q is greater than K, the reaction is backward, and if the Q is smaller than K, then the reaction is forward.