18-Given S- c2H6-225Jmole-k- S- c2H4-220Jmole-k- S- H2-130Jmole-k- then delta S- for the process is C2H4-H2 - C2H6 is

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Standard IX

Mathematics

Dot Product of Two Vectors

18.Given S^° ...

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18.Given S^° c2H6=225Jmole-¹k-¹ S^° c2H4=220Jmole-¹k-¹ S^° H2=130Jmole-¹k-¹ then delta S^° for the process is C2H4+H2 ----- C2H6 is

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Q. Given

S_{C_{2} H_{6}}^{\circ} = 225 J m o l^{- 1} K^{- 1}, S_{C_{2} H_{4}}^{c i r c} = 220 J m o l^{- 1} K^{- 1}, S_{H_{2}}^{\circ} = 130 J m o l^{- 1} K^{- 1}

Then

△ S^{\circ}

for the process

C_{2} H_{4} + H_{2} \to C_{2} H_{6}

is:

K_{P} = 0.04

atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of

C_{2} H_{6}

when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

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

Following gaseous reaction is undergoing in a vessel $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6}; Δ H = - 32.7 K c a l$ . Which will increase the equilibrium concentration of $C_{2} H_{6}$ ?

For the gas phase reaction $C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6},$ $Δ H$ = -32.7kcal carried out in a vessel, the equilibrium concentration of $C_{2} H_{4}$ can be increased by -

Q. For the gas phase reaction,

C_{2} H_{4} + H_{2}

⇌

C_{2} H_{6}

(Δ H = - 32.7 k c a l)

, carried out in a closed vessel. The equilibrium concentration of

C_{2} H_{4}

can be increased by?

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