Question

Use the following data to determine the normal boiling point, in K, of mercury. What assumptions must you make in order to do the calculation?

Hg (l) ?H_t = 0 (by definition) S = 77.4 J/k?mol
Hg (g) ?H_t = 60.78 kJ/mol S = 174.7 J/k?mol
The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ mol and 269.2 J/K mol, respectively. The standard enthalpy of formation and the standard entropy of liquid benzene are 49.04 kJ/mol and 172.8 J/K mol, respectively. Calculate ?H, ?S, and ?G for the process at 25 C.
?H = ......... kJ/mol
?S = .......... J/k?mol
?G = .......... kJ/mol
C_6 H_6(l) ? C_6H_6(g)
Is the reaction spontaneous at 25.00 C?
a. yes
b. no
c. we cannot tell.

Answer 1

The normal boiling point of mercury is determined by setting the Gibbs free energy change for vaporization to zero and solving for the temperature. The enthalpy and entropy changes for the vaporization of benzene are calculated and used to determine the Gibbs free energy change at 25 °C, confirming the process is nonspontaneous at this temperature.

Step-by-step explanation:

To determine the normal boiling point of mercury, we can use the given entropy values (S) and the standard enthalpy of vaporization (ΔHt) at the boiling point. The boiling point can be found where the Gibbs free energy change (ΔG) for the transition from liquid to gas is zero. This occurs when ΔHt - TΔS = 0, where T is the temperature in Kelvin. We need to make the assumption that the given values of ΔHt and S are constant over the temperature range in question, which is generally a good approximation near the boiling point. For benzene, the enthalpy change for the vaporization is Δ2.93 kJ/mol - 49.04 kJ/mol = 33.89 kJ/mol; the entropy change is 269.2 J/K·mol - 172.8 J/K·mol = 96.4 J/K·mol. The Gibbs free energy change at 25 °C (or 298 K) is ΔG = ΔH - TΔS = 33.89 kJ/mol - (298 K)(0.0964 kJ/K·mol) = 33.89 kJ/mol - 28.72 kJ/mol = 5.17 kJ/mol. Since ΔG is positive, the vaporization of benzene at 25 °C is nonspontaneous, so the answer to whether the reaction is spontaneous at 25 °C is 'b. no'.

Answer 2

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Assumptions:

The ideal gas law applies.

The enthalpy change and entropy change values provided are constant over the temperature range of interest.

The enthalpy change and entropy change are independent of pressure.

Now, let's calculate the values:

ΔH = ΔH(g) - ΔH(l)

ΔH = (60.78 kJ/mol) - 0 kJ/mol

ΔH = 60.78 kJ/mol

ΔS = ΔS(g) - ΔS(l)

ΔS = (174.7 J/(K·mol)) - (77.4 J/(K·mol))

ΔS = 97.3 J/(K·mol)

ΔG = ΔH - TΔS

To calculate ΔG at 25°C (298.15 K):

ΔG = (60.78 kJ/mol) - (298.15 K) × (97.3 J/(K·mol)) / (1000 J/kJ)

ΔG = 60.78 kJ/mol - 29.01 kJ/mol

ΔG = 31.77 kJ/mol

Therefore, ΔH = 60.78 kJ/mol, ΔS = 97.3 J/(K·mol), and ΔG = 31.77 kJ/mol for the process of converting liquid benzene to gaseous benzene at 25.00°C.

To determine if the reaction is spontaneous at 25.00°C, we can compare ΔG to zero. If ΔG is negative, the reaction is spontaneous; if ΔG is positive, the reaction is non-spontaneous.

ANSWER IS YES. Since ΔG = 31.77 kJ/mol, which is positive, the reaction is non-spontaneous at 25.00°C.