Question

In the reaction, \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{CaCO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) at \(950^{\circ} \mathrm{C}\) and \(\mathrm{CO}_{2}\) pressure of \(1 \mathrm{~atm}\), the \(\Delta \mathrm{H}\) is found to be 176 \(\mathrm{kJ} /\) mole. Assuming that the volume of the solid phase changes little by comparison with the volume of gas generated, calculate the \(\Delta \mathrm{E}\) for this reaction.

Short answer

The change in internal energy (ΔE) for the reaction \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\) under the given conditions is approximately 175.9 kJ/mol.

Step-by-step solution

Write down the given values

We have: - ΔH = 176 kJ/mol - Temperature, T = 950°C = 1223 K (Converting to Kelvin by adding 273 to temperature in Celsius) - Pressure, P = 1 atm - Gas generated: 1 mole of CO2

Find PΔV using the Ideal Gas Law

We know that PΔV = nRT for 1 mole of a gas, and in this case, the gas generated is CO2. Therefore, PΔV = nRT Where: - P is the pressure - n is the number of moles - R is the gas constant (value is 0.0821 L·atm/K·mol) - T is the temperature in Kelvin Plugging the values: PΔV = (1 atm)(1 mol)(0.0821 L·atm/K·mol)(1223 K) PΔV ≈ 100.53 J/mol (Since 1 kJ = 1000 J, we will convert this to kJ/mol) PΔV ≈ 0.101 kJ/mol

Calculate ΔE using the relationship between ΔH and ΔE

We know that ΔH = ΔE + PΔV We are given the value of ΔH and we have found the value of PΔV. Now we can find the value of ΔE. ΔE = ΔH - PΔV Plugging the values, ΔE = 176 kJ/mol - 0.101 kJ/mol ΔE ≈ 175.9 kJ/mol So, the change in internal energy (ΔE) for this reaction is approximately 175.9 kJ/mol.

Key concepts

Enthalpy (ΔH)

Enthalpy, represented as ΔH, is an important concept in thermodynamics. It represents the total heat content of a system and is often used to measure the heat transferred in a chemical reaction.
Enthalpy change occurs when bonds are broken and formed in a reaction. For the reaction mentioned, the ΔH, or enthalpy change, is 176 kJ/mol. This indicates the heat absorbed or released during the conversion of calcium carbonate (\( \mathrm{CaCO}_{3} \)) to calcium oxide (\( \mathrm{CaO} \)) and carbon dioxide (\( \mathrm{CO}_{2} \)).

• Positive ΔH: endothermic reaction (heat absorbed).
• Negative ΔH: exothermic reaction (heat released).

The given ΔH value of 176 kJ/mol is positive, suggesting this reaction requires energy input to break the bonds.

Internal Energy (ΔE)

Internal Energy, denoted as ΔE, refers to the total energy contained within a system. Unlike enthalpy, it does not include energy expended through pressure-volume work.
The relationship between ΔH and ΔE is given by:\[ ΔH = ΔE + PΔV \]where PΔV represents the work done by the system as it expands or contracts at constant pressure.
In the exercise, ΔE was calculated using this relationship, with ΔE turning out to be approximately 175.9 kJ/mol. This value indicated the change in internal energy when accounting for the minor work done (PΔV) by the released gas, \( \mathrm{CO}_{2} \).

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics, expressed as:\[ PV = nRT \]This law links pressure (P), volume (V), number of moles (n), the ideal gas constant (R), and temperature (T).

• R has a value of 0.0821 L·atm/K·mol.
• In calculations, temperature (T) must be in Kelvin.

In the given problem, this law allowed us to calculate PΔV for carbon dioxide gas formed:\[ PΔV = nRT \]By plugging in the values, we derived PΔV as 0.101 kJ/mol for the given conditions of 950°C (converted to 1223 K) and 1 mole of \( \mathrm{CO}_{2} \) gas at 1 atm.

Pressure-Volume Work

Pressure-volume work accounts for the energy expended when a gas expands or contracts against external pressure. This concept is crucial for understanding energy changes while keeping volume change in mind.
In chemical reactions with gases, the expansion of gas can do work on the surrounding environment. This is expressed mathematically as:\[ W = PΔV \]where W is the work done, P is the pressure, and ΔV is the change in volume.

• This work is often small compared to changes in enthalpy.
• In the exercise, PΔV was roughly 0.101 kJ/mol.

By subtracting this work from ΔH, we accurately determined ΔE for the reaction, refining our understanding of internal energy in these thermodynamic calculations.