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The value of K(p) at 298 K for the react...

The value of `K_(p)` at `298 K` for the reaction
`1/2 N_(2)+ 3/2 H_(2) hArr 2NH_(3)`
is found to be `826.0`, partial pressure being measured atmospheric units. Calculate `DeltaG^(ɵ)` at `298 K`.

Text Solution

Verified by Experts

`triangleG^(0)=-2.303 RT log K_(p)`
`=-2.303 xx 1.98 xx 298 xx log 826`
=-3980 calories.
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Knowledge Check

  • K_(C ) for the reaction (1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) hArr NH_(3(g)) is

    A
    `K_(c )=([NH_(3)])/([N_(2)][H_(2)])`
    B
    `K_(c )=([N_(2)][H_(2)])/([NH_(3)])`
    C
    `K_(c )=([NH_(3)])/([N_(2)]^(1/2)[H_(2)]^(3/2))`
    D
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  • The unit of K_(c) from the reaction N_(2)+H_(2)hArr2NH_(3)+Q is

    A
    `lit^(2)mol^(-2)`
    B
    `mollit^(-1)`
    C
    `mol^(2)lit^(2)`
    D
    `lit mol^(-2)`

  • For the reaction, N_(2)+3H_(2)hArr2NH_(3) , the units of K_(c) "and" K_(p) respectively are:

    A
    `mol^(-2)L^(2) "and" b ar^(-2)`
    B
    `mol^(-2)L^(2) "and" B ar^(-1)`
    C
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    D
    `mol^(-1)L^(-1) "and" b ar^(-1)`

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