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.
1. To calculate the standard molar entropy, Sº, of C2H6 gas, we can use the equation ∆Sº = Sº(products) - Sº(reactants). We know the values of ∆Sº and Sº for C2H2 and H2, so we can solve for the Sº of C2H6.
∆Sº = -232.7 J/mol∙K = [(1)(Sº_C2H6)] - [(1)(200.9 J/mol∙K) + (2)(130.7 J/mol∙K)]
Solving for Sº_C2H6, we get:
Sº_C2H6 = -232.7 + 200.9 + 2(130.7)
Sº_C2H6 = 229.6 J/mol∙K
2. To determine the standard free-energy change, ∆Gº, we use the equation:
∆Gº = ∆Hº - T∆Sº, where T is the temperature in Kelvin.
First, we must find ∆Hº for the reaction. Using the given ∆Hºf values:
∆Hº = ∆Hºf(products) - ∆Hºf(reactants)
∆Hº = [(1)(-84.7 kJ/mol)] - [(1)(226.7 kJ/mol) + (2)(0 kJ/mol)]
∆Hº = -84.7 - 226.7 = -311.4 kJ/mol
Now, we can find ∆Gº using the equation above with T = 298 K:
∆Gº = -311.4 kJ/mol - ((298 K)(-232.7 J/mol∙K)(1 kJ/1000 J))
∆Gº = -311.4 kJ/mol + 69.4 kJ/mol
∆Gº = -242.0 kJ/mol
The negative sign of ∆Gº indicates that the reaction is spontaneous under standard conditions.
3. We can find the equilibrium constant, K, using the equation:
∆Gº = -RT ln(K), where R is the gas constant (8.314 J/mol∙K).
Solving for K:
K = e^(-∆Gº / RT)
K = e^(242000 J/mol / (8.314 J/mol∙K * 298 K))
K = 2.37 × 10^20
4. To find the C≡C bond energy in C2H2, we can use the given bond energies for the other bonds in the reaction and the values of ∆Hºf for the reaction:
∆Hº = Bonds broken - Bonds formed
∆Hº = (1 C≡C + 2 H-H) - (1 C-C + 3 C-H)
We already found ∆Hº as -311.4 kJ/mol. The bond energies for H-H and C-C are given as 436 kJ/mol and 347 kJ/mol, respectively. The bond energy for C-H is given as 414 kJ/mol. We can now write the equation as:
-311.4 kJ/mol = (1 C≡C + 2 * 436) - (347 + 3 * 414)
Now we can solve for the C≡C bond energy:
C≡C = -311.4 + 347 + 3 * 414 - 2 * 436
C≡C = 812 kJ/mol
Thus, the value of the C≡C triple bond energy in C2H2 is 812 kJ/mol.
To calculate the standard molar entropy, Sº, of C2H6 gas:
1. From the given data, find the difference in standard entropies (∆Sº) between the products and reactants:
∆Sº = ∆Sº(products) - ∆Sº(reactants)
2. Substituting the given values:
∆Sº = Sº(C2H6) - (Sº(C2H2) + 2Sº(H2))
3. Rearrange the equation to solve for Sº(C2H6):
Sº(C2H6) = ∆Sº + Sº(C2H2) + 2Sº(H2)
4. Substitute the given values:
Sº(C2H6) = -232.7 J/(mol·K) + 200.9 J/(mol·K) + 2(130.7 J/(mol·K))
5. Solve the equation to find Sº(C2H6).
To calculate the standard free-energy change, ∆Gº, for the reaction:
1. Use the equation ∆Gº = ∆Hº - T∆Sº, where T is the temperature in Kelvin.
2. Substitute the given values:
∆Gº = -84.7 kJ/mol - (298 K * (-232.7 J/(mol·K)/1000))
3. Solve the equation to find ∆Gº.
The sign of ∆Gº indicates the spontaneity of the reaction:
- If ∆Gº < 0, the reaction is spontaneous (exergonic).
- If ∆Gº > 0, the reaction is not spontaneous (endergonic).
- If ∆Gº = 0, the reaction is at equilibrium.
To calculate the equilibrium constant for the reaction at 298 K:
1. Use the equation ∆Gº = -RT ln(K), where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin.
2. Rearrange the equation to solve for ln(K):
ln(K) = -∆Gº / (RT)
3. Substitute the given values and solve the equation to find ln(K).
4. Take the exponential of both sides to find K.
To calculate the C-C (triple bond) bond energy in C2H2:
1. Use the equation ∆Hº = ∑∆H(bonds broken) - ∑∆H(bonds formed), where ∆H(bonds broken) is the sum of bond energies for the bonds broken and ∆H(bonds formed) is the sum of bond energies for the bonds formed.
2. Rearrange the equation to solve for the C-C (triple bond) bond energy:
C-C (triple bond) bond energy = (∑∆H(bonds broken) - ∑∆H(bonds formed)) / (number of C2H2 molecules)
3. Substitute the given bond energies and solve the equation to find the C-C (triple bond) bond energy in kJ/mol.
To calculate the standard molar entropy, Sº, of C2H6 gas, you can use the equation:
ΔSº = ΣSº(products) - ΣSº(reactants)
Given that the reaction is:
C2H2(g) + 2 H2(g) → C2H6(g)
The standard molar entropy values (So) for the substances involved are:
C2H2(g): 200.9 J/(mol∙K)
H2(g): 130.7 J/(mol∙K)
C2H6(g): unknown (to be determined)
You are given the standard entropy change (ΔSº) as -232.7 J/(mol∙K). We can use this information to calculate the standard molar entropy of C2H6.
ΔSº = ΣSº(products) - ΣSº(reactants)
-232.7 J/(mol∙K) = Sº(C2H6) - (Sº(C2H2) + 2 x Sº(H2) )
Plug in the given values:
-232.7 J/(mol∙K) = Sº(C2H6) - (200.9 J/(mol∙K) + 2 x 130.7 J/(mol∙K) )
Solve for Sº(C2H6):
Sº(C2H6) = -232.7 J/(mol∙K) + 200.9 J/(mol∙K) + 2 x 130.7 J/(mol∙K)
Sº(C2H6) ≈ 302.2 J/(mol∙K)
Therefore, the standard molar entropy of C2H6 gas is approximately 302.2 J/(mol∙K).
To calculate the value of the standard free-energy change, ΔGº, for the reaction, you can use the equation:
ΔGº = ΔHº - TΔSº
Given:
ΔHº = ∆Hºf(C2H6) - ∆Hºf(C2H2) - 2 x ∆Hºf(H2) = -84.7 kJ/mol - 226.7 kJ/mol - 2 x 0 kJ/mol = -337.4 kJ/mol (note that the ΔHºf values are provided)
ΔSº = -232.7 J/(mol∙K) (given)
T = temperature = 298 K
Plug the values into the equation:
ΔGº = (-337.4 kJ/mol) - (298 K x (-232.7 J/(mol∙K)) ) *Note: convert -232.7 J/(mol·K) to kJ/(mol·K)
ΔGº = -337.4 kJ/mol - (298 K x (-0.2327 kJ/(mol∙K)) )
ΔGº ≈ -266.1 kJ/mol
The sign of ΔGº indicates the spontaneity of the reaction.
If ΔGº < 0, the reaction is spontaneous in the forward direction.
If ΔGº > 0, the reaction is non-spontaneous in the forward direction.
If ΔGº = 0, the reaction is at equilibrium.
The equilibrium constant (K) can be calculated using the equation:
ΔGº = -RT ln(K)
Given:
ΔGº = -266.1 kJ/mol (from the previous calculation)
R = gas constant = 8.314 J/(mol∙K) (conversion factor needed)
T = 298 K
Convert ΔGº from kJ to J:
ΔGº = -266100 J/mol
Plug the values into the equation:
-266100 J/mol = -8.314 J/(mol∙K) x 298 K x ln(K)
Simplify and solve for ln(K):
ln(K) = -266100 J/mol / (-8.314 J/(mol∙K) x 298 K)
ln(K) ≈ 32.171
Finally, solve for K:
K = e^(ln(K))
K ≈ e^(32.171)
K ≈ 6.43469 x 10^13
Therefore, the value of the equilibrium constant for the reaction at 298 K is approximately 6.43469 x 10^13.
To calculate the value of the C-C (triple bond) bond energy in C2H2 in kJ/mol, you need to calculate the sum of the bond energies broken and formed during the reaction using the bond energy values given.
The reaction is:
C2H2(g) + 2 H2(g) → C2H6(g)
In this reaction, two C-H bonds and one C-C (triple bond) are broken, and six C-H bonds are formed.
Bond energy to break 2 C-H bonds = 2 x 414 kJ/mol = 828 kJ/mol
Bond energy to break 1 C-C (triple bond) = 1 x 611 kJ/mol = 611 kJ/mol
Therefore, the total energy required to break the bonds in C2H2 is 828 kJ/mol + 611 kJ/mol = 1439 kJ/mol.
Hence, the value of the C-C (triple bond) bond energy in C2H2 is approximately 1439 kJ/mol.