Question

Match the List I (equilibria) with the List II (conditions that favors the product formation) and select the correct answer code from the options given below:

List I	List II
(P) \[{H_{2(g)}} + {I_{2(g)}} \rightleftharpoons 2H{I_{(g)}}\]	1. High temperature
(Q) \[2S{O_{2(g)}} + {O_{2(g)}} \rightleftharpoons 2S{O_3}_{(g)}\]	2. Low temperature
(R) \[2N{H_{3(g)}} \rightleftharpoons {N_{2(g)}} + 3{H_2}_{(g)}\]	3. High pressure
	4. Low pressure
	5. independent of pressure

A.\[P - 1,Q - 3,R - 2,4\]
B.\[P - 2,Q - 1,4,R - 1,3\]
C.\[P - 1,5,Q - 2,3,R - 2,4\]
D.\[P - 2,4,Q - 1,5,R - 1,3\]

Answer 1

Hint: In the equilibrium reaction, the concentration of both reactants and products are equal and there will not be any tendency to change its concentration with time. And According to Le Chatelier's principle, if adding more reactants to the given system, the reaction will shift towards the right and increase the concentration of the products.

Complete answer:
In the reaction of hydrogen and iodine, there is a formation of two moles of hydrogen iodide. But it will not depend on the pressure. And in the case of the second reaction it takes place at both low temperature as well as high pressure. Hence, option (A) is incorrect.
The reaction of hydrogen and iodine will not take place at low temperature and the conditions of the remaining two reactions are also wrong. Hence, the option (B) is incorrect.
According to Le Chatelier's principle, in the first reaction, one mole of iodine is reacted with one mole of hydrogen and there is a formation of two moles of hydrogen iodide. This reaction takes place at low temperature and it is independent of pressure.
The reaction of sulphur dioxide and oxygen takes place at both low temperature and high pressure.
And in the third reaction, two moles of ammonia are decomposed into one mole of nitrogen and one mole of hydrogen, and this reaction occurs at low pressure as well as low temperature. Therefore the answer is, \[P - 1,5,Q - 2,3,R - 2,4\]. Hence, option (C) is correct.
According to Le Chatelier's principle, the reaction of hydrogen and iodine does not take place at low temperature and low pressure. And other options are also wrong. Hence, the option (D) is incorrect.
Hence, option (C) is correct.

Note:
We must have to know that the Le Chatelier's principle is also known as equilibrium law and it states that, when as system undergoes a disturbances like change in pressure, temperature and concentration etc, and the system has the response to restore the new equilibrium state. When the temperature increases, the reaction favors the endothermic. And the catalyst will not affect the equilibrium.