Problem 30 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The surface of copper gets tarnished by the formation of copper oxide. \(\mathrm{N}_{2}\) gas was passed to prevent the oxide formation during heating of copper at \(1250 \mathrm{~K}\). However, the \(\mathrm{N}_{2}\) gas contains 1 mole \(\%\) of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: \(2 \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g})\) \(p_{\mathrm{H} 2}\) is the minimum partial pressure of \(\mathrm{H}_{2}\) (in bar) needed to prevent the oxidation at \(1250 \mathrm{~K}\). The value of \(\ln \left(p_{\mathrm{H} 2}\right)\) is (Given: total pressure \(=1\) bar, \(R\) (universal gas constant \()=8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \ln\) \((10)=2.3 . \mathrm{Cu}(\mathrm{s})\) and \(\mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})\) are mutually immiscible. At \(1250 \mathrm{~K}: 2 \mathrm{Cu}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s}) ;\) $$ \Delta G^{\circ}=-78,000 \mathrm{~J} \mathrm{~mol}^{-1} $$ $$ \begin{aligned} &\mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ; \Delta G^{\circ}=-1,78,000 \mathrm{~J} \mathrm{~mol}^{-1}\\\ &\text { ( } G \text { is the Gibbs energy) } \end{aligned} $$

Step-by-Step Solution

Answer

The value of \(\ln(p_{\text{H}_2})\) is -14.6.

Step 1: Understand the Reaction System

We need to analyze the potential oxidation of copper by water vapor during heating. The objective is to determine the hydrogen partial pressure that would prevent this oxidation. Two given reactions illustrate the formation and prevention of copper oxidation by providing the associated Gibbs free energies at 1250 K.

Step 2: Calculate Gibbs Free Energy for Oxidation

Using the given reactions, calculate the Gibbs free energy change (ΔG°) for the oxidation of copper to copper oxide in the presence of water vapor. This involves the reaction: \[ 2 \text{Cu(s)} + \text{H}_2\text{O(g)} \rightarrow \text{Cu}_2\text{O(s)} + \text{H}_2\text{(g)} \] Calculate the overall ΔG° for this using the ΔG° values from decomposition into simplified reactions.

Step 3: Gibbs Free Energy Change Analysis

The reaction of copper with water vapor can be decomposed into: 1. Reaction of copper with oxygen to form Cu₂O \[ 2 \text{Cu} + \frac{1}{2} \text{O}_2 \rightarrow \text{Cu}_2\text{O} \quad \Delta G_1^{\circ} = -78,000 \text{ J/mol} \]2. Reaction of hydrogen with oxygen to form water vapor\[ \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O} \quad \Delta G_2^{\circ} = -178,000 \text{ J/mol} \]Subtract the second equation from the first to find the Gibbs free energy change for the desired oxidation reaction: \[ \Delta G_{\text{oxidation}}^{\circ} = \Delta G_1^{\circ} - \Delta G_2^{\circ} = 100,000 \text{ J/mol} \]

Step 4: Calculate Equilibrium Constant

The equilibrium constant (K) for the reaction can be found using the relationship: \[ \Delta G^{\circ} = -RT \ln(K) \]By setting \( T = 1250 \text{ K} \) and \( R = 8 \text{ J K}^{-1} \text{ mol}^{-1} \), we have: \[ 100,000 = -8 \times 1250 \times \ln(K) \]This simplifies to:\[ \ln(K) = -10 \]

Step 5: Express Reaction Quotient

Using the reaction quotient \( Q \) for the system, we express it in terms of \( p_{\text{H}_2} \) and \( p_{\text{H}_2\text{O}} \) since it is given that the total pressure is 1 bar and 1% water vapor means \( p_{\text{H}_2\text{O}} = 0.01 \):\[ K = \frac{p_{\text{H}_2}}{0.01} \]

Step 6: Solve for Minimum Partial Pressure

Insert the calculated equilibrium constant back to find the minimum partial pressure of hydrogen that prevents oxidation:\[ \frac{p_{\text{H}_2}}{0.01} = e^{-10} \]Thus,\[ p_{\text{H}_2} = 0.01 \times e^{-10} \]

Step 7: Calculate the Natural Logarithm of p_{H2}

From the solution:\[ \ln(p_{\text{H}_2}) = \ln(0.01) - 10 \]\[ \ln(0.01) = -\ln(100) = -2\times 2.3 = -4.6 \]So,\[ \ln(p_{\text{H}_2}) = -4.6 - 10 = -14.6 \]

Step 8: Interpret the Final Result

The final result tells us the value of the natural logarithm of the minimum hydrogen partial pressure needed to prevent copper oxidation at 1250 K. The result \( \ln(p_{\text{H}_2}) = -14.6 \) indicates this minimal condition.

Key Concepts

Gibbs Free Energy

In the world of chemical thermodynamics, one of the central concepts is Gibbs Free Energy (G), which helps us predict whether a reaction will proceed spontaneously. Gibbs Free Energy combines enthalpy (heat content) and entropy (disorder) to give us a comprehensive view of the thermodynamics of a reaction. For a reaction to occur on its own, without external input, the change in Gibbs Free Energy, denoted as \( \Delta G \), must be negative.

• Negative \( \Delta G \) means the reaction is spontaneous.
• A positive \( \Delta G \) means the reaction requires energy input to proceed.
• If \( \Delta G \) equals zero, the system is at equilibrium.

In our exercise, we focus on the oxidation of copper by water vapor to form copper oxide and hydrogen gas. The standard Gibbs Free Energy changes for the steps involved were given, allowing us to calculate \( \Delta G^{\circ} \) for the overall oxidation. This value tells us whether the oxidation is thermodynamically favorable.

Equilibrium Constant

The equilibrium constant \( K \) is a fundamental concept that quantifies the ratio of products to reactants when a reaction reaches equilibrium. It is a measure of the extent of a reaction; how far it goes before reaching stability, where the forward and reverse reactions occur at the same rate.

• A large \( K \) suggests the reaction heavily favors product formation.
• A small \( K \) suggests the reaction favors reactants, with little product formed.
• \( K \) changes with temperature.

In the exercise, \( \Delta G^{\circ} \) is related to \( K \) by the equation \( \Delta G^{\circ} = -RT \ln(K) \). Here, \( R \) is the gas constant and \( T \) is the temperature in Kelvin. Solving the equation for \( K \) helps us determine the behavior of the reactants and products under specific conditions.

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas within a mixture of gases. Each gas in a mixture contributes to the total pressure as if it were alone in the entire volume. The concept is vital in chemistry, particularly in reactions involving gases, as it helps to explain how gases interact without direct contact.

In our scenario, partial pressure is crucial for understanding how water vapor and hydrogen gas – both present as gases – affect the formation of copper oxide. The task is to ensure that the partial pressure of hydrogen is sufficient to stop the oxidation of copper, maintaining it below the threshold necessary for the reaction to proceed spontaneously. Calculating these partial pressures is vital for controlling reaction outcomes.

Oxidation Reaction

An oxidation reaction involves the loss of electrons by a molecule, atom, or ion. Often, it includes the addition of oxygen or the removal of hydrogen. Oxidation is a critical component of many chemical reactions, especially in processes like corrosion, combustion, and cellular respiration.

In the exercise provided, copper undergoes oxidation. The presence of water vapor, serving as an oxidizing agent, causes copper to lose electrons and form copper oxide while producing hydrogen gas. Understanding oxidation reactions like this allows us to predict and manage processes like metal tarnishing and devise methods to prevent or employ them as needed. For copper, the goal is to impede the reaction to keep the metal untarnished.

Reaction Quotient

The reaction quotient \( Q \) is similar to the equilibrium constant and tells us about the ratio of product concentrations to reactant concentrations at any point in time. While \( K \) describes a system at equilibrium, \( Q \) can be used to determine the direction in which a reaction proceeds to reach equilibrium.

• If \( Q < K \), the reaction moves forward to produce more products.
• If \( Q > K \), the reaction moves in reverse to produce more reactants.
• If \( Q = K \), the system is at equilibrium.

In the oxidation scenario, \( Q \) helps us evaluate the current state of the system compared to \( K \), indicating whether more copper will oxidize or if the reaction will reverse. This plays a critical role in determining the required hydrogen partial pressure to prevent further oxidation of copper.