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The decomposition of NH(3) on finely div...

The decomposition of `NH_(3)` on finely divided platinum follows the rate expression
Rate `=(k_(1)[NH_(3)])/(1+k_(2)[NH_(3)])`
it is a first order reaction when concentration of `NH_(3)` is

A

very low

B

very high

C

moderate

D

never

Text Solution

Verified by Experts

The correct Answer is:
A
Rate `=(k_(1)[NH_(3)])/(1+k_(2)[NH_(3)])`
when `[NH_(3)]` is very low
`1+k_(2)[NH_(3)]~~1`
`therefore` Rate `=k_(1)[NH_(3)]`
Thus, the reaction becomes a first order reaction at low concentration.
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The decomposition of NH_3 on the surface of finely divided platinum as catalyst

The rate of decomposition of ammonia is found to depend upon the concentration of NH_(3) according to the equation -(d[NH_(3)])/(dt)=(k_(1)[NH_(3)])/(1+k_(2)[NH_(3)]) What will be the order of reaction when (i) concentration of NH_(3) is very high ? (ii) concentration of NH_(3) is very low ?

Knowledge Check

  • Decomposition of NH_(3) on the surface of tungsten is a reaction of :

    A
    zero order
    B
    first order
    C
    second order
    D
    fractional order.

  • The kinetics of the decomposition of NH_(3) on the tungsten surface follows.

    A
    Zero order
    B
    First order
    C
    Second order
    D
    Third order

  • The decomposition of NH_(3) on platinum surface is zero order reaction the rate of production of H_(2) is (k=2.5xx10^(-4) M s^(-1))

    A
    `2.5xx10^(-4) M s^(-1)`
    B
    `1.25 xx 10^(-4) M s^(-1)`
    C
    `3.75 xx10^(-4) M s^(-1)`
    D
    `5.0 xx10^(-4) M s^(-1)`

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    The decomposition of NH_(3) on platinum surface is zero order reaction. If rate constant (k) is 4 × 10^(-3) Ms^(-1) , how long will it take to reduce the initial concentration of NH_(3) from 0.1 M to 0.064 M.

    The decomposition of NH, on platinum surface is zero order reaction. If rate constant (k) is 4xx10^(-3) Ms^(-1) , how long will it take to reduce the initial concentration of NH_(3) from 0.1 M to 0.064 M.

    Decomposition of ammonia on platinum surface follows the change 2NH_(3)(g) to N_(2)(g) + 3H_(2)(g) If the rate obeys (-d[NH_(3)])/(dt)=(k_(1)[NH_(3)])/(1+k_(2)[NH_(3)]), what will be the order for decomposition of NH_(3) if [NH_(3)] is very very less, K_(1)"and" K_(2) are constant?

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