For a given reaction lnk-100 - 103- T -4then - S0 for the reaction is in calmol -1

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Thermodynamic Processes

For a given r...

Question

For a given reaction
$l n k = \frac{- (100 \times 10^{3})}{T} + 4$
then $Δ S^{0}$ for the reaction is (in $c a l m o l^{- 1}$ )

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K

T

ln K = 3.0 + \frac{2.0 \times 10^{4}}{T}

R = 8.3 J K^{- 1} m o l^{- 1}

Δ H^{o}

Δ S^{o}

l n K \to \frac{1}{T} =

△ H^{0} = - 76.6 K J

Δ S^{0} = 226 J K^{- 1}

8

c a l m o l^{- 1}

x

c a l m o l^{- 1}

x

C O C l_{2} (g)

C O C l_{2} (g) \to C O (g) + C l_{2} (g); Δ H = 19 k c a l / m o l

l n k = 15 - \frac{5000}{T}

(i n K c a l / m o l)

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