# 5.1g of $N{{H}_{4}}SH$ is introduced in 3.0L evacuated flask at ${{327}^{0}}C$. 30% of the solid $N{{H}_{4}}SH$ decomposed to $N{{H}_{3}}$and ${{H}_{2}}S$ as gases. The ${{K}_{p}}$ of the reaction at ${{327}^{0}}C$ is:

(R=$0.082\,Latm\,mo{{l}^{-1}}{{K}^{-1}}$, molar mass of S=$32g\,mo{{l}^{-1}}$, molar mass of N=$14g\,mo{{l}^{-1}}$)

A. $1\times {{10}^{-4}}at{{m}^{2}}$

B. $4.9\times {{10}^{-3}}at{{m}^{2}}$

C. $0.242at{{m}^{2}}$

D. $0.242\times {{10}^{-4}}at{{m}^{2}}$

Answer

Verified

181.8k+ views

**Hint:**To solve this question, find out the individual number of moles for the products as well as the reactants. Then use the ideal gas equation $PV=nRT$ and find out the pressures of the components. Finally, substitute the values in the equation of ${{K}_{p}}$

**Complete step-by-step answer:**

${{K}_{p}}$ is the constant calculated from the partial pressures of a reaction equation. it's wont to express the link between product pressures and reactant pressures. it's a unitless number, although it relates the pressures. A homogeneous equilibrium is one within which everything within the equilibrium mixture is present within the same phase. During this case, to use ${{K}_{p}}$, everything must be a gas. Now, in order to answer the question, let us write the reaction:

$N{{H}_{4}}SH->N{{H}_{3}}+{{H}_{2}}S$, when time=0, then number of moles of $N{{H}_{4}}SH$=$\dfrac{given\,mass}{molar\,mass}$which is equal to $\dfrac{5.1}{51}=0.1$ moles.

Now, after time ${{t}_{0}}$, we have the concentration of $N{{H}_{4}}SH$ as (0.1-x), assuming x as concentra. So, on the product side, as we have 1 mole of each, the concentration of $N{{H}_{3}}$ and ${{H}_{2}}S$ respectively will be ‘x’. But, we have been given that 30% of $N{{H}_{4}}SH$decomposed to the gases. So the value of x will be $\dfrac{30}{100}=0.3$. Now using the ideal gas equation, we have

$PV=nRT$ So,

${{P}_{N{{H}_{3}}}}\times 3=0.03\times 0.082\times 600$, which comes out to be 0.492 atm, and is same for $N{{H}_{3}}\,and\,{{H}_{2}}S$. Now, let us write the expression for ${{K}_{p}}$:

${{K}_{p}}=[{{P}_{N{{H}_{3}}}}][{{P}_{{{H}_{2}}S}}]$

= ${{(0.492)}^{2}}\,at{{m}^{2}}$

**=$0.242\,at{{m}^{2}}$, which gives us option C as the answer.**

**NOTE:**The concentration of $N{{H}_{4}}SH$ was not used in the equation of ${{K}_{p}}$ as it is a solid and it’s concentration is counted as unity. Only concentration of gases can be put in the expression of ${{K}_{p}}$. As there is no reactant in the denominator, the unit is ${{(atm)}^{2}}$, instead of $atm$.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main