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N(2)3H(2)hArr2NH(3) K=4xx10^(6) "at" 298...

`N_(2)3H_(2)hArr2NH_(3) K=4xx10^(6) "at" 298`
`K=41 "at" 400 K`
Which statements is correct?

A

If `N_(2)` is added at equilibrium condition, the equilibrium will shift to the forward direction because according to `II^(nd)` law of thermodynamics the entropy must increases in the direction of spontaneous reaction.

B

The condition for equilibrium is `2DeltaG_(NH_(3))=3DeltaG_(H_(2)+DeltaG_(N_(2))` where `G` gibbs free energy per mole of the gaseous species measured at that partial pressure.

C

Addition of catalyst does not change `K_(P)` but changes `DeltaH`.

D

At `400K` addition of catalyst will increase forward reaction by `2` times while reverse reaction rate will be changed by `1.7` times.

Text Solution

Verified by Experts

The correct Answer is:
B
When nitrogen is added at equilibrium condition, the equilibrium will shift according to Le-chatelier principle at equilibrium `DeltaG=0` and catalyst changes the rate of forward and backward reactions by equal extent. K_(P)` of reaction is a function of temperature only.
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Knowledge Check

  • N_(2) + 3H_(2) hArr 2NH_(3)" "K=4xx 10^(6)"at"298 " "K=41 "at" 400 k Which statements is correct?

    A
    If `N_(2)` is added at equlibrium condition, the equilibrium will shift to the forward direction because according to `II^(nd)` law of thermodynamics the entropy must increases in the direction of spontaneous reaction .
    B
    The condition for equlibrium is `2DeltaG_(NH_(3)) = 3DeltaG_(N_(2)) + DeltaG_(N_(2))` where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure.
    C
    Addition of catalyst does not change `K_(p)` but changes `DeltaH`.
    D
    At 400 K addition of catalyst will increase forward reaction by `2` times while reverse reaction rate will be changed by 1.7 times.

  • The equilibrium constant K_(p) for the reaction, N_(2)(g) +3H_(2)(g) hArr 2NH_(3)(g)" is "1.6xx10^(-4)" (atm)"^(-2)" at "400^(@) . What will be the equilibrium constant at 500^(@) if heat of the reaction in this temperature range is -25.14 kCal ?

    A
    `1.231xx10^(-4) ("atm")^(-2)`
    B
    `1.876xx10^(7) ("atm")^(-2)`
    C
    `1.462xx10^(-5) ("atm")^(-2)`
    D
    `3.462xx10^(-5) ("atm")^(-2)`

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    A
    The reaction is exothermic
    B
    Increase in temperature increases the formation of B
    C
    Increase in pressure increases the formation of A
    D
    Decrease in temperature and increase in pressure shift the equilibrium towards left

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