how do I solve this please? Thermodynamic Quantities for Selected Substances at

Question

how do I solve this please?

Thermodynamic Quantities for Selected Substances at 298.15 K (25 C) Carbon C (s, diamond) C (s, graphite) C2H2 (g) C2H4 (g) C2H6 (B) co (g) CO2 (8) 2.43 5.69 200.8 219.4 229.5 197.9 213.6 2.84 0 1.88 226.7 209.2 52.30 -84.68 110.5 68.11 32.89 -137.2 -393.5 394.4 Hydrogen H2( 8) 0 0 130.58 Oxygen 0 0 205.0 02 (8) H20 (1) 285.83 237.13 69.91 1)Calculate the value of S'm assum and ASunv for the catalytic hydrogenation of acetylene to ethene . C2H2 (g) + H2 (g) C2H4 (g) Is the reaction spontaneous?

Explanation / Answer

Dear Student,

Given

C2H2(g) + H2(g) ----- C2H4(g)

entropy of C2H2(g) = 200.8 J/K-mol

entropy of H2(g) = 130.58 J/K-mol

entropy of C2H4(g) = 219.4 J/K-mol

To find change in entropy in the system,surroundings and universe.

Formula:

S(system) = S(final) - S(intial)

Solution:

S(system) = S(final) - S(intial)

S(system) = 219.4 - (130.58 +200.8)

S(system) = 219.4 - 331.38

S(system) = -112.0 J/K-mol

Therefore , entropy change in the system is -112.0 J/K-mol and - 0.112 KJ/K-mol

To find entropy change in the surroundings we have to first find H which is given by the formula

H = H(fproducts) - H(reactants)

H of C2H4 = 52.30 KJ/mol

H of C2H2 = 226.7 KJ/mol

H of H2 = 0 KJ/mol

H = 52.30 - (226.70 + 0 )

H = 52.30 - 226.70

H = - 174.40 KJ/mol

S(surroundings) = H / T

S(surroundings) = -174.40 / 298.15

S(surroundings) = -0.5849 KJ/K-mol

Therefore, entropy change in the suroundings is - 0.5849 KJ/K-mol

S(universe) = S(system) + S(surroundings)

S(universe) = -0.112 +(-0.5849)

S(universe) = -0.112-0.5849

S(universe) = - 0.6969 KJ/mol

Therefore, entropy change in the universe is - 0.6969KJ/mol.

According to the second law of thermodynamics the entropy of the universe must increase in a spontaneous process. It is possible for the entropy of system to decrease as long as entropy of surroundings increase.

Hence the above reaction is spontaneous reaction.

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