C2H2(g) + 2 H2(g)--> C2H6(g)
Information about the substances involved in the reaction represented above is summarized in the following tables.
Substance/ So (J/mol∙K) /∆Hºf (kJ/mol)
C2H2(g) / 200.9 / 226.7
H2(g) / 130.7 / 0
C2H6(g)/ ?? / -84.7
Bond Bond Energy (kJ/mol)
C-C 347
C=C 611
C-H 414
H-H 436
1.If the value of the standard entropy change, ∆Sº for the reaction is -232.7 joules per mole∙Kelvin, calculate the standard molar entropy, Sº, of C2H6 gas.
2.Calculate the value of the standard free-energy change, ∆Gº, for the
reaction. What does the sign of ∆Gº indicate about the reaction above?
3.Calculate the value of the equilibrium constant for the reaction at 298 K.
4.Calculate the value of the C C(triple bond) bond energy in C2H2 in kJ/mole.
C C= [ C triple bond C]
To answer these questions, we need to use the information provided about the substances involved in the reaction and their standard entropies, standard enthalpies of formation, and bond energies. Let's go step by step.
1. To calculate the standard molar entropy, Sº, of C2H6 gas, we need to use the equation:
ΔSº = ΣnSº(products) - ΣmSº(reactants)
In this case, we only have C2H6 gas as the product. Using the values given in the table, we have:
ΔSº = Sº(C2H6) - (2 * Sº(H2) + Sº(C2H2))
ΔSº = Sº(C2H6) - (2 * 130.7 + 200.9)
Given that ΔSº = -232.7 J/mol∙K, we can rearrange the equation to solve for Sº(C2H6):
Sº(C2H6) = ΔSº + (2 * 130.7 + 200.9)
Sº(C2H6) = -232.7 + (2 * 130.7 + 200.9)
Sº(C2H6) = -232.7 + 392.3
Sº(C2H6) = 159.6 J/mol∙K
2. The standard free-energy change, ΔGº, for the reaction can be calculated using the equation:
ΔGº = ΣnΔGº(products) - ΣmΔGº(reactants)
In this case, we can use the equation:
ΔGº = ΣnΔHºf(products) - ΣmΔHºf(reactants) - TΔSº
Substituting the values from the table into the equation:
ΔGº = ΔHºf(C2H6) - (ΔHºf(C2H2) + 2 * ΔHºf(H2)) - T * ΔSº
ΔGº = -84.7 - (226.7 + 2 * 0) - 298 * (-232.7 / 1000) (converting J to kJ)
ΔGº = -84.7 - (226.7 + 0) - 298 * (-0.2327)
ΔGº = -84.7 - 226.7 + 69.5641
ΔGº = -241.803 kJ/mol
The sign of ΔGº indicates that the reaction is exergonic with a negative ΔGº value, indicating that the reaction is spontaneous under standard conditions.
3. To calculate the value of the equilibrium constant for the reaction at 298 K, we can use the equation:
ΔGº = -RT * ln(K)
ΔGº = -241.803
R = 8.314 J/mol·K (universal gas constant)
T = 298 K
Substituting the values into the equation:
-241.803 = -8.314 * 298 * ln(K)
ln(K) = -241.803 / (-8.314 * 298)
ln(K) = 1.001
K = e^(1.001)
K ≈ 2.719
Therefore, the value of the equilibrium constant (K) for the reaction at 298 K is approximately 2.719.
4. To calculate the value of the C-C (triple bond) bond energy in C2H2, we need to subtract the energies of the individual bonds of C-C and C=C from the energy of the C-C triple bond.
C-C bond energy = C-C triple bond energy - C=C bond energy
C-C bond energy = 611 kJ/mol - 347 kJ/mol
C-C bond energy = 264 kJ/mol
Therefore, the value of the C-C (triple bond) bond energy in C2H2 is 264 kJ/mol.