Q

# Need explanation for: - Equilibrium - JEE Main

An amount of solid $NH_{4}HS$ is placed in a flask already containing ammonia gas at a certain temperature and $0.50\: atm$ pressure. Ammonium hydrogen sulphide decomposes to yield $NH_{3} and H_{2}S$ gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.85 atm. The equilibrium constant for$NH_{4}HS$ decomposition at this temperature is

• Option 1)

$0.30$

• Option 2)

$0.18$

• Option 3)

$0.17$

• Option 4)

$0.11$

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As we learnt in

Relation between pressure and concentration -

$PV=nRT$

$or\:P=\frac{n}{V}RT$

$or\:P=CRT$

$R=0.0831\:bar\:inter/mol\:K$

- wherein

P is pressure in Pa. C is concentration in mol / litre. T is temperature in kelvin

and

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

$aA+bB\rightleftharpoons cC+dD$

$K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}$

$[A],\:[B],\:[C]\:[D]$

are equilibrium concentration

$NH_{4}HS(s)\rightleftharpoons NH_{3}(g)+H_{2}S(g)$

Initial pressure             0                                0.5                    0

At equi.                         0                            0.5 + x                  x

Total pressure = 0.5 + 2x = 0.84

$\therefore$    x = 0.17 atm

$K_{p}=p_{H_{3}}s\times p_{H_{2}}s=(0.5+0.17)(0.17)=0.11\ atm^{2}$

Option 1)

$0.30$

This option is incorrect

Option 2)

$0.18$

This option is incorrect

Option 3)

$0.17$

This option is incorrect

Option 4)

$0.11$

This option is correct

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