Question

Use the data found in the Thermodynamic Values link under Reference Materials to calculate ΔH^o (KJ), ΔS^o (J/K) and ΔG^o(KJ) and ΔG^o from the equation ΔG^o ≈ ΔH^o - TΔS^o at 25.0 ^oC for  

1.  C₂H₅OH _(g)  →   C₂H_{4 (g)} + H₂O _(l)

2.  2 C₄H_{10 (l)} + ___ O_{2 (g)}  →  ___ CO_{2 (g)} + ___ H₂O _(l)    (Do not change a coefficient that is shown)

3. 2 C₄H_{10 (g)} + ___ O_{2 (g)}  →  ___ CO_{2 (g)} + ___ H₂O _(l)    (Do not change a coefficient that is shown)

THERMODYNAMIC VALUES FOR SUBSTANCES AT 298.15 K (25°C) Substance ΔΗ, (kJ/mol) AGE (kJ/mol) (J/mol-K) -705.6 -1669.8 -630.0 -1576.5 28.3 109.3 51.0 Aluminum AL AICI Al.O Barium Baa BaCO Bao Beryllium Весы BcO Be(OH)2 Bromine -1216.3 -553.5 -1137.6 -525.1 9.44 -608.4 -905.8 -579.1 -817.9 50.2 Brad Bra Brig Br2) 111.8 -120.9 30.7 82.4 -102.8 3.14 174.9 80.7 245.3 152.3 198.5 HBO -36.2 -53.2 179.3 145.5 Callum Cap Ca CaCON Сась Cafe Cao Ca(OH) Caso Carbon -1207.1 -795.8 -1219.6 -635.5 -986.2 -1434.0 1128.8 -748.1 -1167.3 -604.2 -898.5 -1321.8 154.8 41.4 92.9 104.6 68.9 39.8 83.4 106.7 718.4 672.9 158.0

Answer 1

1)

C2H5OH) + C2H4(9) + H,00

Given:-

AH (C2H5OH)) = -235.1(kJ/mol)

AH (C2H4(9)) = 52.3(kJ/mol)

AH (H2O) = -285.8(kJ/mol)

Now,
ΔΗΣ ΔΗ(product) –Σ ΔΗ?(reactant)

ΔΗ° = [1(mol) xΔΗ (CH4)+1(mol) xΔΗ (Η2Ο)]-[1(mol) xΔΗ (CHOH)

AH' = [1(mol) 52.3(kJ/mol) +1(mol) * -285.8(kJ/mol)] – [1(mol) -235.1(kJ/mol]

AH° = 1.6(KJ) = 1600(J)

Now,

S" (C2H5OH)) = 282.7(J/mol.K)

sº(C2H4(q)) = 219.4(J/mol.K)

S' (H2O) = 69.9(J/mol.K)

AS" => sº(product) - S"(reactant)

AS0 = [1(mol) x S" (C2H4)+1(mol) x S" (H20)]- 1(mol) x S" (CH3OH)

ASC = [1(mol) x 219.4(J/mol.K)+1(mol) x 69.9(J/mol.K)] – [1(mol) x 282.7(J/mol.K)]

ASO = 6.6(J/K)

T = 25 °C = 298 K

AG = AH-TAS

AG = 1600(J) – 298(K) x 6.6(J/K)

AG" = 1600(J) – 1966.8(J) = -366.8(J) = -0.3668(kJ)

--------------------------------

2)

2C4H10 (1) + 1302(g) + 8C02(g) + 10,0(1)

Given :-

AH (C4H10()) = -147.6(kJ/mol)

AH (O2(g)) = 0(kJ/mol)

AH(CO2(g)) = -393.5(kJ/mol)

AH (H2O) = -285.8(kJ/mol)

Now,
ΔΗΣ ΔΗ(product) –Σ ΔΗ?(reactant)

| AH = [8(mol) x H (CO2)+10(mol) x H (H2O)]-[2(mol)xAH(CH 10)+ 13(mol) ~ AH (02)]

AH = [8(mol)x-393.5(kJ/mol)+10(mol) –285.8(kJ/mol)] – [2(mol) -147.6(kJ/mol) + 13(mol) x 0(kJ/mol)]

AH° = -5710.8(kJ) = -5710800(J)

Now,

Sº (C4H10(0) = 231(J/mol.K)

sº (O2(g)) = 205(J/mol.K)

sº (CO2(g)) = 213.6(J/mol.K)

S' (H2O) = 69.9(J/mol.K)

AS" => sº(product) - S"(reactant)

ASC = [8(mol) x S°(CO2)+10(mol) x Sº(H20)] – [2(mol) x S" (C4H10) + 13(mol) x sº(O2)]

ASC = [8(mol) x 213.6(J/mol.K)+10(mol)x 69.9(J/mol.K)] – [2(mol) x 231(J/mol.K) + 13(mol) x 205(J/mol.K)]

AS" = 1852.8(J/K) + 699(J/K) - [462(J/K) + 2665(J/K)

AS" = -575.2(J/K)

Now,

T = 25 °C = 298 K

AG = AH-TAS

AG° = -5710800(J) - (298(K) * -575.2(J/K))

AG = -5710800(J) - (-171409.6(J))

AGC = -5539390.4() = -5539.3904(kJ)