C2H2(g) + 2 H2(g)--> C2H6(g)
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
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.
Calculate the value of the standard free-energy change, ∆Gº, for the reaction. What does the sign of ∆Gº indicate about the reaction above?
Calculate the value of the equilibrium constant for the reaction at 298 K.
Calculate the value of the C C (triple bond) bond energy in C2H2 in kJ/mole.
The standard molar entropy of C2H6 gas is -232.7 J/mol∙K.
The standard free-energy change, ∆Gº, for the reaction is -112.4 kJ/mol. The sign of ∆Gº indicates that the reaction is exothermic.
The equilibrium constant for the reaction at 298 K is 0.0014.
The C C (triple bond) bond energy in C2H2 is 822 kJ/mol.
To calculate the standard molar entropy, Sº, of C2H6 gas, you can use the equation:
ΔSº = ΣnSº(products) - ΣnSº(reactants)
Given that the ∆Sº for the reaction is -232.7 J/mol∙K, and we know the standard entropies of C2H2, H2, and C2H6, we can substitute the values into the equation:
-232.7 J/mol∙K = (1 mol)(Sº of C2H6) - (1 mol)(Sº of C2H2) - (2 mol)(Sº of H2)
Sº of C2H6 = (1 mol)(Sº of C2H2) + (2 mol)(Sº of H2) - 232.7 J/mol∙K
Next, we need to find the values for Sº of C2H2 and Sº of H2, which we are given.
Sº of C2H2 = 200.9 J/mol∙K
Sº of H2 = 130.7 J/mol∙K
Plugging in the values:
Sº of C2H6 = (1 mol)(200.9 J/mol∙K) + (2 mol)(130.7 J/mol∙K) - 232.7 J/mol∙K
Sº of C2H6 = 462.3 J/mol∙K
Therefore, the standard molar entropy of C2H6 gas is 462.3 J/mol∙K.
To calculate the standard free-energy change, ∆Gº, for the reaction, you can use the equation:
∆Gº = ∆Hº - T∆Sº
Given that ∆Hºf of C2H6 is -84.7 kJ/mol, and we have calculated the value for Sº of C2H6 as 462.3 J/mol∙K, we can substitute the values into the equation. However, we also need to convert the units to match:
∆Gº = (-84.7 kJ/mol) - (298 K)(0.2327 kJ/mol∙K)
∆Gº = -84.7 kJ/mol - 69.2996 kJ/mol
∆Gº = -153.9996 kJ/mol
Therefore, the value of the standard free-energy change, ∆Gº, for the reaction is approximately -154 kJ/mol.
The sign of ∆Gº indicates that the reaction is spontaneous because ∆Gº is negative. A negative ∆Gº indicates that the reaction is exergonic and can occur spontaneously without any additional energy input.
To calculate the value of the equilibrium constant, K, for the reaction at 298 K, you can use the equation:
∆Gº = -RT ln(K)
Where R is the gas constant (8.31 J/mol∙K), and T is the temperature (in Kelvin).
Substituting the values:
-153.9996 kJ/mol = - (8.31 J/mol∙K)(298 K) ln(K)
-153.9996 kJ/mol = - 2468.38 J ln(K)
Now, we need to rearrange the equation to solve for K:
ln(K) = (-153.9996 kJ/mol) / (-2468.38 J)
ln(K) = 156.6591
K = e^156.6591
Therefore, the value of the equilibrium constant, K, for the reaction at 298 K is approximately e^156.6591.
To calculate the value of the C-C (triple bond) bond energy in C2H2 in kJ/mol, you can use the bond energy values given in the question:
C=C bond energy = 611 kJ/mol
C-H bond energy = 414 kJ/mol
Since the reaction involves the formation of C2H6 from C2H2, we can use the difference in bond energies to calculate the C-C (triple bond) bond energy in C2H2.
C2H2(g) + 2 H2(g) --> C2H6(g)
Bond energy change = Σ (bonds broken) - Σ (bonds formed)
In this reaction, one C-C bond and four C-H bonds are formed, while one C=C bond and two H-H bonds are broken.
Bond energy change = (1 mol)(611 kJ/mol) + (4 mol)(414 kJ/mol) - (1 mol)(C-C bond energy) - (2 mol)(H-H bond energy)
Since we want to calculate the C-C (triple bond) bond energy in C2H2, we only need the C-C bond energy. So, we rearrange the equation:
C-C bond energy = (1 mol)(611 kJ/mol) + (4 mol)(414 kJ/mol) - (2 mol)(H-H bond energy)
C-C bond energy = (1 mol)(611 kJ/mol) + (4 mol)(414 kJ/mol) - (2 mol)(436 kJ/mol)
C-C bond energy = 611 kJ/mol + 1656 kJ/mol - 872 kJ/mol
C-C bond energy = 1395 kJ/mol
Therefore, the C-C (triple bond) bond energy in C2H2 is 1395 kJ/mol.
To calculate the standard molar entropy, Sº, of C2H6 gas, you can use the equation:
∆Sº = ΣSº(products) - ΣSº(reactants)
Given the data, the entropy values for the substances involved are:
Sº(C2H2) = 200.9 J/mol∙K
Sº(H2) = 130.7 J/mol∙K
Sº(C2H6) = unknown
Since the reactants are C2H2 and H2, and the products are C2H6, the equation becomes:
∆Sº = Sº(C2H6) - [Sº(C2H2) + 2Sº(H2)]
Substituting the known values:
-232.7 J/mol∙K = Sº(C2H6) - [200.9 J/mol∙K + 2(130.7 J/mol∙K)]
Solving for Sº(C2H6):
Sº(C2H6) = -232.7 J/mol∙K + 200.9 J/mol∙K + 2(130.7 J/mol∙K)
Sº(C2H6) = 79.8 J/mol∙K
Therefore, the standard molar entropy of C2H6 gas is 79.8 J/mol∙K.
To calculate the standard free-energy change, ∆Gº, for the reaction, you can use the equation:
∆Gº = ∆Hº - T∆Sº
Given the values for ∆Hºf and ∆Sº, you can calculate:
∆Hº = Σ∆Hºf(products) - Σ∆Hºf(reactants)
T = temperature in Kelvin (298 K in this case)
Substituting the known values:
∆Gº = (-84.7 kJ/mol) - (298 K * (-232.7 J/mol∙K / 1000 J/kJ))
∆Gº = -84.7 kJ/mol + (298 K * (-0.2327 kJ/mol))
∆Gº = -84.7 kJ/mol - 69.2876 kJ/mol
∆Gº = -153.9876 kJ/mol
The sign of ∆Gº indicates that the reaction is spontaneous. Negative ∆Gº values indicate exothermic, spontaneous reactions.
To calculate the equilibrium constant, K, for the reaction at 298 K, you can use the equation:
∆Gº = -RT ln(K)
Given the value of ∆Gº and the gas constant, R = 8.314 J/(mol∙K), you can calculate:
-153.9876 kJ/mol = -8.314 J/(mol∙K) * 298 K * ln(K)
Simplifying:
-153.9876 kJ/mol = -2471.97214 J * ln(K)
ln(K) = -153.9876 kJ/mol / -2471.97214 J
ln(K) = 62.3
Taking the exponential of both sides:
K = e^(62.3)
K ≈ 3.62 x 10^26
Finally, to calculate the value of the C-C (triple bond) bond energy in C2H2, you can subtract the energies of the other bonds involved from the enthalpy change (∆Hº) for the reaction:
∆Hº = Σ∆H(bonds broken) - Σ∆H(bonds formed)
Given the bond energy values:
C-C = 347 kJ/mol
C=C = 611 kJ/mol
C-H = 414 kJ/mol
H-H = 436 kJ/mol
The C2H2 molecule contains one C-C triple bond. All other bonds (C=C and C-H) are formed in the reaction. Therefore:
∆Hº = (∆H(triple bond broken)) - 2(∆H(C-H formed))
∆Hº = (C-C bond energy) - 2(C-H bond energy)
Substituting the bond energy values:
∆Hº = (347 kJ/mol) - 2(414 kJ/mol)
∆Hº = -481 kJ/mol
Therefore, the C-C (triple bond) bond energy in C2H2 is approximately 481 kJ/mol.