For a reaction with H 24 kcal mol and S 0 07 kcalmol1K1,calculate the ...

1 We know that G H T S Given H 24 kCalmol S 007 kCalmolK T temperature in kelvi ....

For a reaction with Î”Hâˆ˜=âˆ’24 kcal/mol and Î”Sâˆ˜=0.07 kcalmolâˆ’1Kâˆ’1, calculate the equilibrium constant at: (1.) 30...

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Chemistry

For a reaction with Î”Hâˆ˜=âˆ’24 kcal/mol and Î”Sâˆ˜=0.07 kcalmolâˆ’1Kâˆ’1,calculate the equilibrium constant at: (1.) 30 âˆ˜C and (2.) 150âˆ˜C.

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(1) We know that Î”Gâˆ˜ = Î”Hâˆ˜ - T Î”Sâˆ˜

Given Î”Hâˆ˜ = -24 kCal/mol =

Â Â Â Â Â Â Â Î”Sâˆ˜ = 0.07 kCal/mol-K

Â Â Â Â Â Â Â Â T = temperature in kelvin = 30 oC = 30+273 = 303 K

Plug the values we get Î”Gâˆ˜ = -24 - ( 303x0.07)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = -45.21 Kcal/mol

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = -45.21x10³ Cal /mol

We know that Î”Gâˆ˜ = -RT ln K

Where R = gas constant = 2 Cal/mol-K

T = temperature = 303 K

K = Equilibrium constant = ?

Plug the values we get ln K = -Î”Gâˆ˜/ (RT)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = -(-45.21x10³) / ( 2x303)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 74.6

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â K = e^74.6

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.51x10³²

Simillarly do the remaining

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For a reaction with Î”Hâˆ˜=âˆ’24 kcal/mol and Î”Sâˆ˜=0.07 kcalmolâˆ’1Kâˆ’1, calculate the equilibrium constant at: (1.) 30...

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80.2K

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Chemistry

For a reaction with Î”Hâˆ˜=âˆ’24 kcal/mol and Î”Sâˆ˜=0.07 kcalmolâˆ’1Kâˆ’1,calculate the equilibrium constant at: (1.) 30 âˆ˜C and (2.) 150âˆ˜C.

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