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For the reversible reaction N(2)(g)+3H...

For the reversible reaction
`N_(2)(g)+3H_(2)(g)hAr2NH_(3)(g)`
at `500^(@)C` the value of `K_(p)` is `1.44xx10^(-5)` when partial pressure is measured in atmosphere. The corresponding value of `K_(e)` with concentration in mol/L is

A

`(1.44xx10^(-5))/((0.082xx500)^(-2))`

B

`(1.44xx10^(-5))/((8.314xx773)^(-2))`

C

`(1.44xx10^(-5))/((0.082xx773^(2)))`

D

`(1.44xx10^(-5))/((0.082xx773)^(-2))`

Text Solution

Verified by Experts

The correct Answer is:
D
`N_(2)+3H_(2)hArr2NH_(3)`
`Deltan=2-4=-2K_(c)=(K_(p))/((RT)^(Deltan))=(1.44xx10^(-5))/((0.082xx773)^(-2)),`
(R in L. atm. `K^(-1)mol^(-1)`)
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For the reversible reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) at 500^(@)C , the value of K_(p) is 1.44xx10^(-5) when the partial pressure is measured in atmophere. The corresponding value of K_(c) with concentration in mol L^(-1) is

For the reversible reaction, N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) At 500^(@)C , the value of Kp is 1.44 xx 10^(-5) when partial pressure is measured in atmosphere. The corresponding value of Kc with concentration in mol//L is:

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