Home
Class 11
CHEMISTRY
For the equilibrium N(2)+3H(2) hArr 2N...

For the equilibrium
`N_(2)+3H_(2) hArr 2NH_(3)` at 298K
and 5 atm. Pressure, increasing pressure to 50 atm, will ( here K is the equilibrium constant )

A

decrease the yield of `NH_(3)` as well as the value of K

B

increase the yield of `NH_(3)` as well as the value of K

C

increase the yield of `NH_(3)`, but will not change the value of K

D

increase the yield of `NH_(3)` , but will decrease the value of K

Text Solution

Verified by Experts

The correct Answer is:
C
Equilibrium constant, K depends only upon temperature. Increasing pressure will increase the yield of `NH_(3)` by shifting the equilibrium in the forward direction but will not affect the value of K. It is clear from the following explanation. Let equilibrium concentration of reactants and products are
`[NH_(3)]=.^(n)NH_(2)//V,[N_(2)]=nN_(2)//V`
`[H_(2)]=nH_(2)//V`
where V is the volume of the container.
`K=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=([nNH_(3)//V]^(2))/([nN_(2)//V][nH_(2)//V]^(3))`
`=((nNH_(3))^(2)V^(2))/((nN_(2))(nH_(2)))`
Increasing pressure increases `.^(n)NH_(3)` but also decrases V. This explains as to why K does not change with change in pressure.
Promotional Banner

Topper's Solved these Questions

  • PHYSICAL AND CHEMICAL EQUILIBRIA

    DINESH PUBLICATION|Exercise REASON ASSERTION TYPE MCQs|10 Videos

  • P BLOCK ELEMENTS (GROUP 13 AND 14 )

    DINESH PUBLICATION|Exercise Straight obj.|17 Videos

  • RATES OF REACTIONS AND CHEMICAL KINETICS

    DINESH PUBLICATION|Exercise Ultimate Preparatory Package|29 Videos

Similar Questions

Explore conceptually related problems

For the equilibrium H_(2) O (1) hArr H_(2) O (g) at 1 atm 298 K

On decomposition of NH_(4)HS , the following equilibrium is estabilished: NH_(4) HS(s)hArrNH_(3)(g) + H_(2)S(g) If the total pressure is P atm, then the equilibrium constant K_(p) is equal to

The equilibrium constant for the reversible reaction N_(2)+3H_(2) hArr 2NH_(3) is K and for the reaction (1)/(2) N_(2)+(3)/(2) H_(2) hArr NH_(3) , the equilibrium constant is K',.K and K' will be related as

Ammonium hydrogen sulphide dissociates according to the equation : NH_4 HS(s) hArr NH_3(g)+H_2S(g) The total pressure at equilibrium at 400 K is found to be 1 atm.The equilibrium constant K_p of the reaction is :

For the reaction A_(2)(g) + 2B_(2)hArr2C_(2)(g) the partial pressure of A_(2) and B_(2) at equilibrium are 0.80 atm and 0.40 atm respectively.The pressure of the system is 2.80 atm. The equilibrium constant K_(p) will be

When the reaction, 2NO_(2)(g) hArr N_(2)O_(4)(g) reaches equilibrium at 298 K . The partial pressure of NO_(2) and N_(2)O_(4) are 0.2 Kpa and 0.4 Kpa , respectively. What is the equilibrium constant K_(p) of the above reaction at 298 K ?

For the reaction N_(2)O_(4)(g)hArr2NO_(2)(g) , the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant K_(p) will be

In the reaction, PCl_(5)hArrPCl_(3)+Cl_(2) , the amounts of PCl_(5) , PCl_(3) and Cl_(2) , 2 moles each at equilibrium and the total pressure is 3 atm . Find the equilibrium constant K_(p) .

Knowledge Check

  • For the equilibrium H_(2) O (1) hArr H_(2) O (g) at 1 atm 298 K

    A
    standard free energy change is equal to zero `(Delta G^(@) = 0)`
    B
    free energy change is less than zero `(Delta G lt 0)`
    C
    standard free energy change is less than zero `(Delta G^(@) lt 0)`
    D
    standard free energy change is more than zero `(Delta G^(@) gt 0)`

  • On decomposition of NH_(4)HS , the following equilibrium is estabilished: NH_(4) HS(s)hArrNH_(3)(g) + H_(2)S(g) If the total pressure is P atm, then the equilibrium constant K_(p) is equal to

    A
    P atm
    B
    `P^(2) atm^(2)`
    C
    `P^(2) // 4 atm^(2)`
    D
    2P atm

  • The equilibrium constant for the reversible reaction N_(2)+3H_(2) hArr 2NH_(3) is K and for the reaction (1)/(2) N_(2)+(3)/(2) H_(2) hArr NH_(3) , the equilibrium constant is K',.K and K' will be related as

    A
    `K=K'`
    B
    `K'=sqrt(K)`
    C
    `K=sqrt(K')`
    D
    `KxxK'=1`