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The rate of decomposition of ammonia is ...

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 ?

Text Solution

AI Generated Solution

To determine the order of the reaction for the decomposition of ammonia under two different concentration conditions, we will analyze the given rate equation: \[ -\frac{d[NH_3]}{dt} = \frac{k_1 [NH_3]}{1 + k_2 [NH_3]} \] ### Step 1: Analyze the case when the concentration of \( NH_3 \) is very high. ...
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The rate of a certain reaction depends on concentration according to equation -(d[A])/(dt)=(k_(1)[A])/(1+k_(2)[A]) What will be the order of reaction when (i) concentration is very high (ii) very low ?

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Knowledge Check

  • 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

  • Shape of NH_(3) is very similar to :

    A
    `CH_(4)`
    B
    `CH_(3)^(-)`
    C
    `BH_(3)`
    D
    `CH_(3)^(+)`

  • The shape of NH_(3) is very similar to that of :

    A
    `BH_(3)`
    B
    `CH_(3)^(-)`
    C
    `CH_(3)^(+)`
    D
    `SO_(4)^(2-)`

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    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?

    The decompoistion of ammonia on platinum surface follows the change 2NH_(3) rarr N_(2) + 3H_(2) (a) What does (-d[NH_(3)])/(dt) denote? (b) What does (d[N_(2)])/(dt) and (d[H_(2)])/(dt) denote? ( c) If the decompoistion is zero order then what are the rates of Production of N_(2) and H_(2) if k = 2.5 xx 10^(-4) M s^(-1) ? If the rates obeys - (d[NH_(3)])/(d t) = (k_(1)[NH_(3)])/(1+k_(2)[NH_(3)]) , what will be the order for decompoistion of NH_(3) , if (i) [NH_(3)] is every less and (ii) [NH_(3)] is very high? ( k_(1) and k_(2) are constants)

    For the reaction 2NH_3 to N_2 +3H_2 , if -(d[NH_3])/(dt)=k_1[NH_3] , (d[N_2])/(dt)=k_2[NH_3] , (d[H_2])/(dt)=k_3[NH_3] then the relation between k_1,k_2 and k_3 is

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