Question

The units of \[{K_c}\] for the decomposition of potassium nitrate is same as the units of \[{K_c}\] for the:
A. Decomposition of sulphur trioxide
B. Decomposition of hydrogen iodide
C. Decomposition of ammonium nitrite
D. Formation of nitrogen dioxide from nitric oxide and oxygen

Answer 1

Hint: Decomposition is the process of breakdown of a compound in various fragments and equilibrium constant is the reaction quotient at chemical equilibrium. These two things will give you the answer. You need to know the formula to find the equilibrium constant.

Complete step by step answer:
First, What is decomposition? The answer is that Decomposition of a compound means breaking elements into two or more chemical entities. It is also called chemical breakdown. It is the opposite of chemical synthesis because in chemical synthesis two more entities join to form a compound and decomposition is breaking down of compounds into fragments. And the chemical reaction involved is called decomposition reaction.

Kc is the equilibrium constant of a chemical reaction. It is the value of reaction quotient at chemical equilibrium. It is a state when the dynamic chemical system has reached a level after which it has no tendency towards further change and after sufficient time has elapsed. Stability constants, formation constants, binding constants, association constants and dissociation constants are all types of equilibrium constants. At equilibrium, the rate of forward reaction is equal to rate of backward reaction. Therefore \[{K_c}\] is calculated in moles per litre as given below for chemical reaction

 \[Kc = \dfrac{{Kf}}{{Kb}} = \dfrac{{[C][D]}}{{[A][B]}}\]
Now, decomposition of potassium nitrate is as given below

The \[{K_c}\] will be
 \[{K_c} = \dfrac{{{{[KN{O_2}]}^2} . {[O_2]}}}{{{{[KN{O_3}]}^2}}}\]
The decomposition of sulphur trioxide and its \[{K_c}\] value is as given below

  \[{K_c} = \dfrac{{{{[S{O_2}]}^2} . {[O_2]}}}{{{{[S{O_3}]}^2}}}\]
The decomposition of hydrogen iodide and its \[{K_c}\] value is as given below

 \[{K_c} = \dfrac{{{{[{H_2}]}^{1/2}} . {{[{I_2}]}^{1/2}}}}{{{{[HI]}^{}}}}\]
The decomposition of ammonium nitrite and its \[{K_c}\] value is as given below

 \[{K_c} = \dfrac{{{{[{N_2}]}_{}} .{{[{H_2}O]}^2}}}{{{{[N{H_4}N{O_2}]}_{}}}}\]
The fourth option is invalid because it is a chemical synthesis reaction and \[{K_c}\] of synthesis reaction cannot be compared with decomposition reaction.
From the equations we can observe that the Kc value of sulphur trioxide will be similar to that of potassium nitrate because of similarity in powers in both the equations.

Therefore our answer is option A.

Note:
The equilibrium constant values are different for different reactions and also vary for decomposition and synthesis reactions.