The value of - G- for the reaction at 298 K is

The correct option is B 17.92 Δ H^⊖ = Δ H^⊖_ P Δ H^⊖_ R ; (for 1 mol) = [ 1575.0 kJ mol^ 1 3 / 2 × 241.8 ] [ 2021.0 kJ mol^ 1 ] ...

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Standard XII

Chemistry

Nernst Equation

The value of ...

Question

The value of $Δ G^{⊖}$ for the reaction at $298 K$ is

120

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17.92

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- 180

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Solution

The correct option is B $17.92$
$Δ H^{⊖} = Δ H_{P}^{⊖} - Δ H_{R}^{⊖}; (f o r 1 m o l)$
$= [- 1575.0 k J m o l^{- 1} - \frac{3}{2} \times 241.8] - [- 2021.0 k J m o l^{- 1}]$
$= + 83.3 k J m o l^{- 1}$
For 1 Kg $C a S O_{4} \cdot 2 H_{2} O$
Numebr of moles $= \frac{1000}{172}$
$= 5.81$
$∴$ Heat change for $5.81 m o l o f C a S O_{4} \cdot 2 H_{2} O = 5.81 \times 83.3 k J m o l^{- 1}$
$= 484 k J m o l^{- 1}$
$Δ G = Δ H - T Δ S$
$= 17.92 k J$ ........ $(Δ S = S_{P} - S_{R})$

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The value of ΔG⊖ for the reaction at 298K is

The value of $Δ G^{⊖}$ for the reaction at $298 K$ is