Consider the reaction N_{2}(g)+3H_2(g)\rightleftharpoons 2N\! \! H_3(g) . The equilibrium constant of the above reaction is K_{p} . If pure ammonia is left to dissociate , the partial pressure of ammonia at equilibrium is given by ( Assume thatP_{NH_{3}}< < p_{total}  at equilibrium )

  • Option 1)

    \frac{3^{\frac{3}{2}}K_{p}^{\frac{1}{2}}P^{2}}{16}

  • Option 2)

    \frac{3^{\frac{3}{2}}K_{p}^{\frac{1}{2}}P^{2}}{4}

  • Option 3)

    \frac{K_{p}^{\frac{1}{2}}P^{2}}{4}

  • Option 4)

    \frac{K_{p}^{\frac{1}{2}}P^{2}}{16}

Answers (1)
A admin

 

Law of Chemical equilibrium -

A+B\rightleftharpoons C+D

where  A & B are the reactants, C & D are the product in balanced chemical equations.

- wherein

K_{c}=\frac{[C]\:[D]}{[A]\:[B]}

Kc is the equilibrium constant.

 

 

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 

 

 

 

Relation between Kp and Kc -

K_{p}=K_{c}(RT)^{\bigtriangleup n}

while calculating the value of Kp , pressure should be expressed in bar.

1\:bar=10^{5}pa

- wherein

\bigtriangleup n = (number of moles of gaseous products) - (number of moles gaseous reaction)

as we learned

2NH_{3}\rightleftharpoons N_{2}+3H_{2}\, \, \, ;K^{1}_{p}=\frac{1}{K_{p}}

Ptotal=P=P_{N_{2}}+P_{H_{2}}+P_{NH_{3}}\approx P_{N_{2}}+P_{H_{2}}

P_{N_{2}}= \frac{P}{4}\, \, ;P_{H_{2}}= \frac{3P}{4}

\frac{1}{K_{P}}= \frac{P_{N_{2}}\left ( P_{H_{3}} \right )^{3}}{\left ( P_{NH_{3}} \right )^{2}}=\frac{\left ( \frac{P}{4} \right )\left ( \frac{3P}{4} \right )^{3}}{\left ( P_{NH_{3}} \right )^{2}}

{\left ( P_{NH_{3}} \right )^{2}}= \frac{3^{3}P^{4}}{4^{4}}K_{p}

P_{NH_{3}}= \frac{3^{3/2}K_{p}^{1/2}P^{2}}{16}

 

 


Option 1)

\frac{3^{\frac{3}{2}}K_{p}^{\frac{1}{2}}P^{2}}{16}

Option 2)

\frac{3^{\frac{3}{2}}K_{p}^{\frac{1}{2}}P^{2}}{4}

Option 3)

\frac{K_{p}^{\frac{1}{2}}P^{2}}{4}

Option 4)

\frac{K_{p}^{\frac{1}{2}}P^{2}}{16}

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