K c for the reaction- F e 2- Zn - Zn 2- F e is 1023- The standard free energy change isA- -1-36 K cal mol -1B- -13-6 K cal mol -1C- -13-6 K cal mol -1D- -1-36 K cal mol -1

The correct option is C 13.6 K cal mol^ 1 Δ G^0 = 2.303 RT log K = 2.303 × 2 × 298 × log 10^23 = 13,580 cal mol^ 1 ...

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Standard XII

Chemistry

Enthalpy of Solution

K c for the r...

Question

$K_{c}$ for the reaction, $F e^{2 +} + Z n \to Z n^{2 +} + F e$ is $10^{23}$ , The standard free energy change is

$+ 1.36 K c a l m o l^{- 1}$

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$- 1.36 K c a l m o l^{- 1}$

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$+ 13.6 K c a l m o l^{- 1}$

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$- 13.6 K c a l m o l^{- 1}$

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Solution

The correct option is D
$- 13.6 K c a l m o l^{- 1}$

$Δ G^{0} = - 2.303 R T - l o g K = - 2.303 \times 2 \times 298 \times l o g 10^{23} = - 13, 580 c a l m o l^{- 1}$

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Kc for the reaction, Fe2++Zn→Zn2++Fe is 1023, The standard free energy change is

The correct option is D −13.6 K cal mol−1 ΔG0=−2.303 RT−log K=−2.303×2×298×log1023=−13,580 cal mol−1

$K_{c}$ for the reaction, $F e^{2 +} + Z n \to Z n^{2 +} + F e$ is $10^{23}$ , The standard free energy change is

The correct option is D
$- 13.6 K c a l m o l^{- 1}$

$Δ G^{0} = - 2.303 R T - l o g K = - 2.303 \times 2 \times 298 \times l o g 10^{23} = - 13, 580 c a l m o l^{- 1}$