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Standard XII

Chemistry

Elevation in Boiling Point

Calculate the...

Question

Calculate the free energy change for the dissociation of potassium nitrate in water at 298K. (given: delta H = 34 KJ/mol, delta S = 0.116KJ/K/mol)

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Solution

Free energy =G = H - (TS)
=G=34x10³-(298x0.116x10³)
G=34000-(34568)
G=568joule

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Similar questions

Q. The Born Haber cycle below represents the energy changes occurring at 298K when

K H

is formed from its elements
v :

{Δ H}_{a t o m i s a t i o n}

K = 90 k J / m o l

w :

{Δ H}_{i o n i s a t i o n}

K = 418 k J / m o l

x :

{Δ H}_{d i s s o c i a t i o n}

H = 436 k J / m o l

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{Δ H}_{e l e c t r o n a f f i n i t y}

H = 78 k J / m o l

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{Δ H}_{l a t t i c e}

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Q. The Gibbs free energy change for the reaction for which

Δ H = - 393.4 k J

and

Δ S = - 2.9 {J K}^{- 1} {m o l}^{- 1}

at 298K is:

The Born Haber cycle below represents the energy changes occurring at 298K when

K H

is formed from its elements
v :

{Δ H}_{a t o m i s a t i o n}

K = 90 k J / m o l

w :

{Δ H}_{i o n i s a t i o n}

K = 418 k J / m o l

x :

{Δ H}_{d i s s o c i a t i o n}

H = 436 k J / m o l

y :

{Δ H}_{e l e c t r o n a f f i n i t y}

H = 78 k J / m o l

z :

{Δ H}_{l a t t i c e}

K H = 710 k J / m o l

On complete reaction with water,

0.1 g

K H

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Q. The Born Haber cycle below represents the energy changes occurring at 298K when

K H

is formed from its elements
v :

{Δ H}_{a t o m i s a t i o n}

K = 90 k J / m o l

w :

{Δ H}_{i o n i s a t i o n}

K = 418 k J / m o l

x :

{Δ H}_{d i s s o c i a t i o n}

H = 436 k J / m o l

y :

{Δ H}_{e l e c t r o n a f f i n i t y}

H = 78 k J / m o l

z :

{Δ H}_{l a t t i c e}

K H = 710 k J / m o l

Out of v to y the most endothermic is___.

Q. Calculate the dissociation constant of

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N H_{3} + H^{\oplus} ⇌ \oplus N H_{4}

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Δ H^{⊝} = - 52.2 k J m o l^{- 1}, Δ S^{⊝} = 1.67 J K^{- 1} m o l^{- 1}

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Calculate the free energy change for the dissociation of potassium nitrate in water at 298K. (given: delta H = 34 KJ/mol, delta S = 0.116KJ/K/mol)

Free energy =G = H - (TS) =G=34x10³-(298x0.116x10³) G=34000-(34568) G=568joule

Free energy =G = H - (TS)
=G=34x10³-(298x0.116x10³)
G=34000-(34568)
G=568joule