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  • Can someone help me with this, - 5.1 g is introduced in 3.0 L evacuated Flask at 3270C . 30% of solid decomposed to as gases . th - Equilibrium - JEE Main

 5.1 g NH_{4}SH is introduced in 3.0 L evacuated Flask at 3270C . 30% of solid NH_{4}SH decomposed  to NH_{3}andH_{2}Sas gases . the Kp of the reaction at 3270C is (R= 0.082 L atm mol _1 K-1, Molar mass of S=32  g mol-1 , molr mass of N= 14 g mol-1 )

  • Option 1)

    0.242 atm2

  • Option 2)

    4.9 X10-3 atm2

  • Option 3)

    0.242 X10-4 atm2

  • Option 4)

    1 X10-4 atm2

Answers (1)

best_answer

 

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

 

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 

 

  As we have learned in equilibrium

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

n = \frac{5.1}{51} = 0.1 mole

Initial             0.1                           0                       0    

               .1(-1-\alpha)                      .1\alpha                      .1\alpha

\alpha = 30 % = 0.3

So number of moles at equilibrium

.1(1-.3)                .1 x .3                      .1 x .3

=0.07                       0.3                         =0.3

Now if we use -

PV= nRT at equilibrium

Ptotal x 3 lit = (.03 + 0.03) x ).082 x 600

Ptotal = 0.984 atm

At equilibrium ,

P_{NH_{3}} = P_{H_{2}S} = \frac{P_{total}}{2} = 0.492

So, K_{P} = P_{NH_{3}} . P_{H_{2}S} = (0.492) (0.492)

K_{P} = 0.242 atm^{2}

 


Option 1)

0.242 atm2

Option 2)

4.9 X10-3 atm2

Option 3)

0.242 X10-4 atm2

Option 4)

1 X10-4 atm2

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