Question

What would be the value of $\Delta {{n}_{g}}$ the reaction?
$N{{H}_{4}}C{{l}_{\left( s \right)}}\rightleftharpoons N{{H}_{3\left( g \right)}}+HC{{l}_{\left( g \right)}}$
(a) 1
(b) 0.5
(c) 1.5
(d) 2

Answer 1

Hint:From the lower classes we are studying a reaction system and about the various reactions taking place in the system. The given question is about the equilibrium system and we could calculate the value of $\Delta {{n}_{g}}$using the equation, $\Delta {{n}_{g}}={{n}_{p\left( g \right)}}-{{n}_{r\left( g \right)}}$

Complete step-by-step answer:In the question, it is asked what will be the value of the change in moles of gases taking place during the given reaction.
The reaction which is of our consideration is a decomposition reaction in which the ammonium chloride is getting converted to ammonia and hydrochloric acid. Another important thing that should be noted in the equation is the reaction arrow given.
The arrow given is double-headed, the two-way arrow, so we can say that the given reaction is in the equilibrium state.
We are very much familiar with the term equilibrium state. It is a state of a reaction in which the forward and the backward reaction will take place at the same rate, hence we cannot observe any physical change in the reaction mixture.
Here the given system is in an equilibrium state and we have to find the change in the number of moles of gases substance in the reaction, ie the difference between the number of moles of a gaseous substance in the product side and the number of moles of a gaseous substance in the reactant side.
Hence we can calculate $\Delta {{n}_{g}}$using the equation given: $\Delta {{n}_{g}}={{n}_{p\left( g \right)}}-{{n}_{r\left( g \right)}}$
In the reaction equation given, there are no reactants that exist in gaseous form, hence ${{n}_{r\left( g \right)}}=0$
On the product side, both the products formed are in the gaseous state, hence ${{n}_{p\left( g \right)}}=2$
Substitute these values in the above equation and we get,$\Delta {{n}_{g}}=2-0=2$

Hence the correct answer for the above-given question is option (d).

Note:There are two types of equilibrium systems, homogeneous equilibrium, and heterogeneous equilibrium.
In a homogeneous equilibrium system all the products formed and the reactants taking part in the reaction will be of the same phase.
In a heterogeneous equilibrium system, the products and the reactants will be in different phases, the above-given system is an example of heterogeneous equilibrium as ammonium chloride is in solid-state and the products formed in the gaseous state.
Always give a value zero for substance in the liquid state and solid-state for doing calculations for the gaseous system and we should also know the physical states of the substances as in some cases they may not provide the physical state.