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Question: Solid ammonium carbonate dissociates to give ammonia and carbon dioxide as follows: \( N{H_2}COON...

Solid ammonium carbonate dissociates to give ammonia and carbon dioxide as follows:
NH2COONH4(s)2NH3(g)+CO2(g)N{H_2}COON{H_4}(s) \to 2N{H_3}(g) + C{O_2}(g)
At equilibrium, ammonia is added such that partial pressure of NH3N{H_3} now equals the original total pressure. Calculate the ratio of the total pressure now to the original total pressure.
(A) 3127\dfrac{{31}}{{27}}
(B) 6040\dfrac{{60}}{{40}}
(C) 319\dfrac{{31}}{9}
(D) 6227\dfrac{{62}}{{27}}

Explanation

Solution

Hint : Partial Pressure is defined as if a container filled with more than one gas, each gas exerts pressure. The pressure of any gas within the container is called its partial pressure. The pressure that is exerted by one among the mixture of gasses if it occupies the same volume on its own is known as Partial pressure.

Complete Step By Step Answer:
In this question it is given that sodium ammonium carbonate which is NH2COONH4N{H_2}COON{H_4} dissociates to give ammonia and carbon dioxide. Now at equilibrium ammonia is added such that the partial pressure of NH3N{H_3} equals the total pressure so pNH3=pT{p_{N{H_3}}} = {p_T} at equilibrium
So let us write the equation firstly for solving
NH2COONH4(s)2NH3(g)+CO2(g)N{H_2}COON{H_4}(s) \to 2N{H_3}(g) + C{O_2}(g)

Products2NH3(g)2N{H_3}(g)CO2(g)C{O_2}(g)
Now let the initial pressure2p2ppp
And the final value of pressure will be3p3ppp'

As it is given in the question that at equilibrium ammonia is added such that the partial pressure of NH3N{H_3} equals the total pressure so pNH3=pT{p_{N{H_3}}} = {p_T} at equilibrium and initially pT=2p+p=3p{p_T} = 2p + p = 3p
As solid does not contribute in pressure so the pressure will only because of ammonia and carbon dioxide
Now the value of kp{k_p} will be kp=(pNH3)2(pCO2){k_p} = {({p_{N{H_3}}})^2}({p_{C{O_2}}})
Now we know the value of partial pressure of ammonia and carbon dioxide which is p
So putting the values of partial pressure in it kp=(2p)2(p){k_p} = {(2p)^2}(p) which is kp=4p3{k_p} = 4{p^3} it is our original total pressure
And for final pressure at equilibrium after the addition of ammonia kp=(3p)2(p){k_p} = {(3p)^2}(p') now comparing both the constants to gain the value of pp'
So it will become kp=4p3=(2p)2(p){k_p} = 4{p^3} = {(2p)^2}(p')
Then value of p=4p9p' = \dfrac{{4p}}{9}
Now the final total pressure will be pT=3p+4p9{p_T} = 3p + \dfrac{{4p}}{9} which is pT=31p9{p_T} = \dfrac{{31p}}{9}
Now taking ratios pT(new)pT(old)=31p93p\dfrac{{{p_{T(new)}}}}{{{p_{T(old)}}}} = \dfrac{{\dfrac{{31p}}{9}}}{{3p}} now the ratio will become pT(new)pT(old)=3127\dfrac{{{p_{T(new)}}}}{{{p_{T(old)}}}} = \dfrac{{31}}{{27}}
Which is our (A) option.

Note :
Initially we took the pressure of carbon dioxide and ammonia pressure in term of p but after the addition of more ammonia the pressure of carbon dioxide will also change so we will not take it as p as before we had taken now its value will be new and we have to find it so that we can get our new pressure.