Question

For the reaction CuS(s) + H2(g) H2S(g) + Cu(s) ΔG°f (CuS) = −53.6 kJ/mol ΔG°f (H2S) = −33.6 kJ/mol ΔH°f (CuS) = −53.1 kJ/mol ΔH°f (H2S) = − 20.6 kJ/mol a. Calculate ΔG° and ΔH° at 298 K and 1 atm pressure. b. Will this reaction proceed spontaneously at 298 K and 1 atm pressure? c. Calculate the equilibrium constant for this reaction at 298 K. d. Calculate ΔS° at 298 K and 1 atm pressure. e. Calculate ΔG at 798 K and 1 atm pressure (assume ΔS° and ΔH° do not change with temperature). f. Calculate the equilibrium constant at 798 K and 1 atm pressure.

Answer 1

a.

To calculate G^o (reaction)

G^o (reaction) = (sum of standard change in free energies of formation of products) - (sum of standard change in free energies of formation of reactants)

G^o (reaction) = a_iG^of_products - b_iG^of_reactants

Also standard change in free energies of formation for elements is zero.

Therefore, G^o (reaction) = G^of(H₂S) - G^of (CuS)

Substituting the given values, G^o (reaction) = -33.6 - (-53.6) = +20.0 kJ/mol

Therefore G^o (reaction) = +20.0 kJ/mol

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To calculate H^o (reaction)

H^o (reaction) = a_iH^of_products - b_iH^of_reactants

Also standard change in enthalpies of formation for elements is zero.

Therefore, H^o (reaction) = H^of(H₂S) - H^of (CuS)

Substituting the given values, H^o (reaction) = -20.6 - (-53.1) = +32.5 kJ/mol

Therefore, H^o (reaction) = +32.5 kJ/mol

Answer for a. is G^o (reaction) = +20.0 kJ/mol and H^o (reaction) = +32.5 kJ/mol

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b.

298K and 1 atm pressure are standard conditions and at standard conditions change in free energy G of reaction is nothing but standard change in free energy G^o of reaction.

For a reaction to be spontaneous change in free energy must be negative but we see that

G (reaction) = G^o (reaction) = +20.0 kJ/mol , is positive at 298K and 1 atm pressure.

Hence, the reaction will NOT proceed spontaneously at 298K and 1 atm pressure.

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c.

Relation between standard change in free energy of reaction G^o (reaction) and equilibrium constant K of the reaction is given by equation, G^o (reaction) = -RTlnK

R = universal gas constant = 8.314 J K^-1mol^-1 ; T = absolute temperature = 298 K (given)

We have calculated G^o (reaction) to be +20.0 kJ/mol = 20*10³ J/mol

Substituting the values in equation we get,

G^o (reaction) = -RTlnK

20*10³ = - 8.314 * 298* lnK

lnK = -8.0724      but lnK = 2.303 log K          ( since lnx = 2.303logx)

Therefore, 2.303 log K = -8.0724          or       logK = -8.0724 / 2.303 = -3.5052

logK = -3.5052 = -3 -0.5052 = -3 - 0.5052 - 1 +1 = (-3-1) + (1-0.5052) = - 4 + 0.4948

logK = - 4 + 0.4948

K = 10^-4 * antilog 0.4948 = 10^-4 *3.125

Therefore equilibrium constant K = 3.125 * 10^-4 at 298K

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d.

Standard change in entropy S^o can be calculated using the relation,

G^o = H^o - TS^o

So, S^o = (H^o - G^o ) / T

We have, G^o = +20.0 kJ/mol, H^o = +32.5 kJ/mol and T = 298 K

Substituting the values we get, S^o = (32.5 - 20) / 298 = 0.04195 kJ mol^-1K^-1 = + 41.95 J mol^-1K^-1

S^o = + 41.95 J mol^-1K^-1   at 298K.

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e.

Change in free energy of reaction G is given by equation,

G = H - TS

Given, T = 798 K,

Assuming H and S are same as H^o and S^o

H = H^o = +32.5 kJ/mol = +32.5 * 10³ J/mol and   S = S^o = + 41.95 J mol^-1K^-1

Substituting the values we get,

G = H - TS = (32.5 * 10³) - (798 * 41.95 ) = - 976.1 J/mol

Therefore, at 798 K, G = - 976.1 J/mol = 0.976 kJ/mol

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f.

Given, T = 798 K

Equilibrium constant K can be calculated using the equation, G^o (reaction) = -RTlnK

G^o (reaction) = +20.0 kJ/mol  = 20*10³ J/mol    ( as calculated under question a)

R = universal gas constant = 8.314 J K^-1mol^-1 ; T = absolute temperature = 798 K (given)

Substituting the values we get,

G^o (reaction) = -RTlnK

20*10³ = - 8.314 * 798 * lnK

20*10³ = - 8.314 * 798 * 2.303 log K               (since lnx = 2.303logx)

log K = - (20*10³) / (8.314 * 798 * 2.303) = -1.3089

log K = -1.3089 = -1 - 0.3089 = -1 -0.3089 +1 -1 = (-1 -1) + (+1 - 0.3089) = - 2 + 0.6911

log K = - 2 + 0.6911

K = 10^-2 * antilog 0.6911 = 10^-2 *4.910

Therefore equilibrium constant K = 4.91 * 10^-2   at 798 K