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Damini Manda
Asked: 2022-11-07T22:04:37+05:30 In:

For the water gas reaction:
`C(s) +H_(2)O(g) hArr CO(g) +H_(2)(g)`
the standard Gibbs enegry for the reaction at `1000K` is `-8.1 kJ mol^(-1)`. Calculate its equilibriu constant.

For the water gas reaction:
`C(s) +H_(2)O(g) hArr CO(g) +H_(2)(g)`
the standard Gibbs enegry for the reaction at `1000K` is `-8.1 kJ mol^(-1)`. Calculate its equilibriu constant.
Parabola
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  1. We know that:
    `Delta_(r)G^(Theta) = – 2.303 RT log K`
    or `logK = (-Delta_(r)G)/(2.303 RT)`
    `Delta_(r)G^(Theta) =- 8.1 kJ mol^(-1), T = 1000K`,
    `R = 8.314 xx 10^(-3) kJ mol^(-1)K^(-1)`
    `:. Log K =- (8.1)/(2.303 xx 8.314 xx 10^(-3) xx 1000) = 0.423`
    or `K = 2.64`

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Naval Gill
Asked: 2022-11-02T16:16:41+05:30 In:

For the water gas reaction,
`C(s) +H_(2)O(g) hArr CO(g)+H_(2)(g)`
the standard Gobbs free energy of reaction (at `1000K)` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.

For the water gas reaction,
`C(s) +H_(2)O(g) hArr CO(g)+H_(2)(g)`
the standard Gobbs free energy of reaction (at `1000K)` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.
Parabola
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  1. We know, `K – antilog ((-DeltaG^(Theta))/(2.3030 RT)) …(i)`
    Given that, `DeltaG^(Theta) =- 8.1 kJ mol^(-1)`
    `R = 8.314 xx 10^(-3) k J K^(-1)mol^(-1)`
    `T = 1000K`
    Substituting these values in equation (i), we get
    `K = antilog [(+(8.1))/(2.303 xx8.314xx 10^(-3)xx1000)]`
    `K = 2.65`

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Naval Bal
Asked: 2022-11-01T23:58:58+05:30 In:

For the water gas reaction:
`C(s) +H_(2)O(g) hArr CO(g) +H_(2)(g)`
the standard Gibbs enegry for the reaction at `1000K` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.

For the water gas reaction:
`C(s) +H_(2)O(g) hArr CO(g) +H_(2)(g)`
the standard Gibbs enegry for the reaction at `1000K` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.
Parabola
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  1. `-DeltaG^(@)=2.303RTlogK_(C)`
    `+8.1xx10^(3)=2.303xx8.314xx1000 logK_(C)`
    `K_(C)=2.648mol litre^(-1)`

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