For the equilibirium- P Cl 5 g- P Cl 3 g- Cl 2 g K c -1-8- 10 -7Calculate - G o for the reaction R-8-314J K -1 mol -1

We know that Δ G ^ o = 2.303RTlog K _ c 2.303× 8.314× 298log (1.8× 10 ^ 7 ) =38484J mol ^ 1 =38.484kJ mol ^ 1 ...

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Gibbs Free Energy & Spontaneity

For the equil...

Question

For the equilibirium,
$P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)$
$K_{c} = 1.8 \times 10^{- 7}$
Calculate $Δ G^{o}$ for the reaction $(R = 8.314 J K^{- 1} {m o l}^{- 1})$

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Solution

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P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)

K_{c} = 1.8 \times 10^{- 7} .

Δ G^{0}

P C l_{5} (g) \Leftrightarrow P C l_{3} (g) + C l_{2} (g), K_{c}

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g),

K_{c} ?

2.0 \times 10^{- 7}

300

(Δ H^{0} = 28.4 k J m o l^{- 1})

P C l_{5 (g)} ⇌ P C l_{3 (g)} + C l_{2 (g)}

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g),

K_{c} .

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