One mole of solid Cr2O3 at 2500 K is dissolved in a large volume of a liquid Raoultian solution of Al2O3 and Cr2O3, in which X_Cr2O3 = 0.2 and which is also at 2500 K. Calculate th... One mole of solid Cr2O3 at 2500 K is dissolved in a large volume of a liquid Raoultian solution of Al2O3 and Cr2O3, in which X_Cr2O3 = 0.2 and which is also at 2500 K. Calculate the changes in enthalpy and entropy caused by the addition. The normal melting temperature of Cr2O3 is 2538 K, and it can be assumed that ΔS_m,Al2O3 = ΔS_m,Cr2O3.
Understand the Problem
The question involves thermodynamic calculations related to the dissolution of Cr2O3 in a solution. It requires calculating the changes in enthalpy and entropy at a specific temperature based on provided conditions.
Answer
$\Delta S = \Delta S_m + 13.38 \, \text{J/(mol K)}$; ΔH depends on additional data.
Answer for screen readers
The change in entropy is given by:
$$ \Delta S = \Delta S_m + 13.38 , \text{J/(mol K)} $$
The change in enthalpy, without specific values for enthalpy changes, cannot be calculated precisely.
Steps to Solve
- Identify the variables and known values
We start with the following known values:
- Temperature, $T = 2500 , \text{K}$
- Mole fraction of $Cr_2O_3$, $X_{Cr_2O_3} = 0.2$
- Normal melting temperature of $Cr_2O_3$, $T_m = 2538 , \text{K}$
- Calculate the change in entropy, ΔS
The formula for the change in entropy when dissolving a solid is given by:
$$ \Delta S = \Delta S_m - R \ln(X_{Cr_2O_3}) $$
Where:
- $\Delta S_m$ is the molar entropy at melting point (for both $Al_2O_3$ and $Cr_2O_3$, assumed equal).
- $R$ is the gas constant, approximately $8.314 , \text{J/(mol K)}$.
Since we don’t have the value of $\Delta S_m$, we will continue with the calculation generically.
- Calculate ΔS with the formula
Substituting the known values into the equation:
$$ \Delta S = \Delta S_m - 8.314 \ln(0.2) $$
Calculating $\ln(0.2)$:
$$ \ln(0.2) \approx -1.609 $$
Thus, we can express ΔS as:
$$ \Delta S \approx \Delta S_m + 8.314 \times 1.609 $$
- Calculate change in enthalpy, ΔH
Assuming that the enthalpy change can be related to the change in temperature from melting, we can express the change in enthalpy as:
$$ \Delta H = \Delta H_m - \Delta H_{fusion} $$
Typically, since the details of $\Delta H_m$ and $\Delta H_{fusion}$ are not given in the problem, we note that $\Delta H$ will depend on $\Delta S$ through Gibbs free energy relation. Without numerical values, we conclude that $\Delta H$ would require more specific data.
- Present final expressions for ΔS and ΔH
Where ΔS has been fully expressed and ΔH awaits specific numeric input for $\Delta H_m$ and $\Delta H_{fusion}$.
The change in entropy is given by:
$$ \Delta S = \Delta S_m + 13.38 , \text{J/(mol K)} $$
The change in enthalpy, without specific values for enthalpy changes, cannot be calculated precisely.
More Information
In thermodynamics, the change in entropy is a crucial factor for assessing the spontaneity of a reaction. The relation between entropy and enthalpy can help determine the feasibility of processes at given temperatures.
Tips
- Forgetting to convert units correctly when working with gas constant (R).
- Neglecting to apply logarithm properties correctly when evaluating $\ln(X)$.
- Not recognizing that the entropy values must be provided or assumed for calculation.
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