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In a catalytic experiment involving Habe...

In a catalytic experiment involving Haber's process, `N_(2)(g)+3" H"_(2)(g)to2" NH"_(3)(g)`, the rate of reaction was measured as : Rate `=(d[NH_(3)])/(dt)=2.0xx10^(-4)"M s"^(-1)`. If there were no side reactions, what was the rate of reaction expressed in terms of `N_(2)` ?

A

`1xx10^(-4)"M s"^(-1)`

B

`4xx10^(-4)"M s"^(-1)`

C

`5xx10^(-3)"M s"^(-1)`

D

`1xx10^(-3)"M s"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
Rate `= (1)/(2)(d[NH_(3)])/(dt) = -(1)/(3)(d[h_(2)])/(dt) = -(d[N_(2)])/(dt)`
`therefore `-(d[N_(2)])/(dt) = (1)/(2) (d[NH_(3)])/(dt) = (1)/(2) xx(2 xx 10^(-4) M s^(-1))`
`" " = Ms^(-1)`
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In a catalyst experiment involving the Haber process N_(2) + 3H_(2) rarr 2NH_(3) , the rate of reaction was measured as Rate = (Delta[NH_(3)])/(Delta t) = 2.0 xx 10^(-4) mol L^(-1) s^(-1) What is the rate of reaction expressed in terms of (a) N_(2) (b) H_(2) ?

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Knowledge Check

  • For the reaction, N_(2)(g)+3H_(2)(g) to 2NH_(3)(g)

    A
    `DeltaH=DeltaE`
    B
    `DeltaHgtDeltaE`
    C
    `DeltaH lt DeltaE`
    D
    None of these

  • In a catalytic conversion of N_(2) to NH_(3) by Haber's process, the rate of reaction was expressed as change in the concentration of ammonia per time is 40 xx 10^(-3) mol L^(-1) s^(-1) . If there are no side reaction, the rate of the reaction as expressed in terms of hydrogen is (in mol L^(-1) s^(-1) )

    A
    `60 xx 10^(-3)`
    B
    `20 xx 10^(-3)`
    C
    `1.200`
    D
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    A
    `60 xx 10^(-3)`
    B
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    C
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