Problem 29 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the reaction, \(2 \mathrm{CO}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2} ; \Delta H=-560 \mathrm{~kJ}\). Two moles of \(\mathrm{CO}\) and one mole of \(\mathrm{O}_{2}\) are taken in a container of volume \(1 \mathrm{~L}\). They completely form two moles of \(\mathrm{CO}_{2}\), the gases deviate appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to \(40 \mathrm{~atm}\), find the magnitude (absolute value) of \(\Delta U\) at \(500 \mathrm{~K}\).(1 \(\mathrm{L}\) atm \(=0.1 \mathrm{~kJ})\)

Step-by-Step Solution

Answer

\(\Delta U = -563 \text{ kJ}\).

Step 1: Identify the Reaction and Given Data

The given reaction is \(2 \text{CO} + \text{O}_2 \longrightarrow 2 \text{CO}_2\) with \(\Delta H = -560 \text{kJ}\). The initial and final pressures are \(70 \text{ atm}\) and \(40 \text{ atm}\), respectively, and the temperature is \(500 \text{ K}\). Note that 1 \(\text{L atm}\) is equivalent to 0.1 \(\text{kJ}\).

Step 2: Understand the Relationship Between \(\Delta H\) and \(\Delta U\)

For reactions involving gas in non-ideal conditions, we need to consider both \(\Delta H\) and work done by the system. The relationship \(\Delta H = \Delta U + P\Delta V\) holds, where \(P\Delta V\) is the work done (in terms of kJ).

Step 3: Calculate \(\Delta n_g\) (Change in Moles of Gas)

Calculate the change in the number of moles of gas (\(\Delta n_g\)) for the reaction. The number of moles of gas decreases from \((2+1) = 3\) moles of reactants to \(2\) moles of products in the reaction, so \(\Delta n_g = 2 - 3 = -1\).

Step 4: Calculate \(P\Delta V\) Using Van Der Waals Equation

Using the initial (\(70 \text{ atm}\)) and final (\(40 \text{ atm}\)) pressures and the change in moles, find \(P\Delta V = (P_f - P_i) \cdot \Delta n_g = (40 \text{ atm} - 70 \text{ atm})(-1) \text{ mole} \cdot 1 \text{ L} = 30 \text{ L atm} = 30 \times 0.1 \text{ kJ} = 3 \text{ kJ}\).

Step 5: Calculate \(\Delta U\) Using \(\Delta H\) and \(P\Delta V\)

The formula is \(\Delta U = \Delta H - P\Delta V\). Substitute the given \(\Delta H = -560 \text{ kJ}\) and calculated \(P\Delta V = 3 \text{ kJ}\) to find \(\Delta U = -560 \text{ kJ} - 3 \text{ kJ} = -563 \text{ kJ}\).

Key Concepts

Enthalpy Change in Reactions

Enthalpy change, denoted by \( \Delta H \), is an essential concept in understanding energy changes during chemical reactions. Essentially, it represents the heat change at constant pressure. When a chemical reaction occurs, bonds between atoms are broken and formed, leading to either absorption or release of energy.

• If \( \Delta H \) is negative, the reaction is exothermic, meaning it releases heat to the surroundings.
• If \( \Delta H \) is positive, the reaction is endothermic, indicating heat absorption from the environment.

In the given reaction \( 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \), the enthalpy change \( \Delta H \) is \(-560 \text{kJ} \). This signifies an exothermic reaction, where 560 kJ of energy is released each time 2 moles of carbon monoxide react with oxygen to form 2 moles of carbon dioxide.

Internal Energy and Work Done

Internal energy, symbolized by \( \Delta U \), is another fundamental concept that describes the total energy within a system. It includes all forms of kinetic and potential energy contained within the particles. During chemical reactions, the change in internal energy is linked to the work done by or on the system. For reactions at constant volume, \( \Delta U \) directly relates to the heat exchanged, but at constant pressure with gas products or reactants, we also account for pressure-volume work. This is expressed as:\[ \Delta U = \Delta H - P\Delta V \]In the context of our exercise, the work done \( P\Delta V \) is connected to the change in pressure and volume, calculated using the change in moles of gas and the pressure difference. Breaking this down:

• Initial pressure \( P_i = 70 \text{ atm} \)
• Final pressure \( P_f = 40 \text{ atm} \)
• \( \Delta n_g = -1 \text{ mole} \), change in moles of gas

Work done is therefore \((P_f - P_i) \cdot \Delta n_g = 3 \text{ kJ}\). This value helps determine the internal energy change as \( \Delta U = -560 \text{ kJ} - 3 \text{ kJ} = -563 \text{ kJ} \).

Chemical Reactions and Non-Ideal Gases

Chemical reactions involve the transformation of reactants into products, influenced by various factors such as temperature, pressure, and especially for gases, their departure from ideal behavior.In ideal gas theory, it is assumed that gas molecules do not interact and occupy no space. However, real gases often deviate from this ideal due to molecular attractions and finite volume. The Van der Waals equation provides a more accurate model by incorporating these factors:\[ \left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT \]where \(a\) and \(b\) are constants specific to each gas, \(V_m\) is the molar volume, and \(R\) is the universal gas constant. In our given exercise, it's important to recognize that the reaction conditions involve non-ideal gases. This affects how we calculate the work done by the reaction and enthalpy changes. In practical scenarios like this, considering non-ideal behavior helps to accurately determine \( \Delta U \), as the prediction from merely ideal assumptions might lead to noticeable errors. Embracing these complexities ensures a more precise understanding of how real gases behave in chemical reactions.