Calculate Kp for the reaction- - Go-100 kJ-mol-1 at 25oC- C2H4g-H2g- C2H6g

We know at equilibrium, Δ G _ P ^ 0 = RTln K _ P ⇒ 100× 10 ^ 3 = 8.314× 298× ln K _ P ⇒ ln K _ P = 100× 10 ^ 3 8.314× 298 =40.362 ∴ #16 ...

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Calorimetry

Calculate K...

Question

Calculate $K_{p}$ for the reaction, $Δ G^{o} = - 100 k J . m o l^{- 1}$ at $25^{o}$ C.
$C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)$

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C_{2} H_{6} (g) ⇌ C_{2} H_{4} (g) + H_{2} (g)

K_{p} = 0.05

Δ G^{o}

627^{o} C

C_{2} H_{6} (g) ⇌ C_{2} H_{4} (g) + H_{2} (g); K_{p} = 0.05 atm

Δ G^{\circ}

627^{\circ} C

(Given : ln (0.05) \approx - 3; R = 8.3 {J K}^{- 1} {mol}^{- 1})

C_{2} H_{6} (g) ⇌ C_{2} H_{4} (g) + H_{2} (g); K_{p} = 0.05 atm

Δ G^{\circ}

627^{\circ} C

(Given : ln (0.05) \approx - 3; R = 8.3 {J K}^{- 1} {mol}^{- 1})

490^{o} C

K_{p}

45.9

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

Δ G^{o}

H_{2} (g) + C_{2} H_{4} (g) \to C_{2} H_{6} (g)

H - H = 103, C - H = 99, C - C = 80

C = C = 145

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

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