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Calculate \(K_{P}\) for the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \quad \Delta G^{\circ}=2.60 \mathrm{~kJ} / \mathrm{mol} $$

Short Answer

Expert verified
\(K_{P} \approx 0.3509\).

Step by step solution

01

Understand the Relationship between Gibbs Free Energy and Equilibrium Constant

The relationship between Gibbs free energy change, \(\Delta G^{\circ}\), and the equilibrium constant, \(K_{P}\), is given by the formula: \(\Delta G^{\circ} = -RT \ln K_{P}\), where \(R\) is the universal gas constant (8.314 J/mol·K) and \(T\) is the temperature in Kelvin.
02

Convert Given Data to Appropriate Units

The temperature is given as \(25^{\circ}C\), which needs to be converted to Kelvin: \(T = 25 + 273.15 = 298.15\, \text{K}\). The \(\Delta G^{\circ}\) is given in \(\text{kJ/mol}\), convert it to \(\text{J/mol}\) by multiplying by 1000, resulting in \(\Delta G^{\circ} = 2600\, \text{J/mol}\).
03

Rearrange the Formula to Solve for \(K_{P}\)

Rearrange the equation \(\Delta G^{\circ} = -RT \ln K_{P}\) to find \(K_{P}\): \( \ln K_{P} = -\frac{\Delta G^{\circ}}{RT} \).
04

Substitute Values into the Rearranged Equation

Substitute \(\Delta G^{\circ} = 2600\, \text{J/mol}\), \(R = 8.314\, \text{J/mol·K}\), and \(T = 298.15\, \text{K}\) into the equation: \( \ln K_{P} = -\frac{2600}{8.314 \times 298.15}\).
05

Calculate \(\ln K_{P}\)

Using the values and the formula, calculate: \( \ln K_{P} = -\frac{2600}{8.314 \times 298.15} \approx -1.0487\).
06

Solve for \(K_{P}\)

To find \(K_{P}\), use the exponential function: \(K_{P} = e^{-1.0487} \approx 0.3509\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs Free Energy, denoted as \( \Delta G \), is a critical concept in chemistry that determines the spontaneity of chemical reactions. It combines enthalpy (\( H \)), entropy (\( S \)), and temperature (\( T \)) into a single value, described by the equation:\[ \Delta G = \Delta H - T\Delta S\]If \( \Delta G \) is negative, the process is spontaneous. Conversely, if positive, the reaction is non-spontaneous. Gibbs Free Energy is fundamental when calculating equilibrium constants because the equation\[\Delta G^{\circ} = -RT \ln K \]links these values directly. Here, \( \Delta G^{\circ} \) represents the standard Gibbs Free Energy change, \( R \) is the gas constant, and \( K \) is the equilibrium constant. This relationship helps predict how far a reaction proceeds before reaching equilibrium, making Gibbs Free Energy a powerful tool for chemists.
Reaction Quotient
The Reaction Quotient, \( Q \), provides a snapshot of a reaction's current state compared to its eventual equilibrium state. It is calculated using the same expression as the equilibrium constant (\( K \)), but involves the reactants and products at any given moment, not just at equilibrium. In a reaction:\[ aA + bB \rightleftharpoons cC + dD \]The expression for \( Q \) is:\[ Q = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]Here, the concentrations or partial pressures of A, B, C, and D are used. By comparing \( Q \) with \( K \), we can determine the direction a reaction proceeds:
  • If \( Q < K \),the reaction will shift right (towards products) to reach equilibrium.
  • If \( Q > K \),the reaction will shift left (towards reactants).
  • If \( Q = K \),the system is already at equilibrium.
Thus, the reaction quotient is an essential tool for chemists to track how reactions proceed within a system.
Chemical Equilibrium
Chemical Equilibrium occurs when the forward and reverse reactions in a chemical system happen at the same rate, resulting in constant concentrations of reactants and products over time. At this state, the system is said to be dynamic because reactions continue to proceed, but no net change is observed. The equilibrium constant,\( K \), quantifies this balance and is calculated using the same expression as the reaction quotient, but exclusively using equilibrium concentrations or pressures.The key point about equilibrium is that it is not about having equal concentrations,but rather a constant ratio governed by the underlying thermodynamics. Factors such as temperature, pressure, and concentration changes can influence an equilibrium state,but not the value of \( K \). Understanding equilibrium helps chemists control conditions to optimize reactions,whether it be maximizing yield or minimizing unwanted compounds.
Thermodynamics in Chemistry
Thermodynamics in Chemistry covers the principles that govern the energy changes in chemical processes. It is built on three foundational laws describing energy conservation, how energy disperses, and the conditions under which systems reach equilibrium. Thermodynamics helps us understand when reactions are favorable, predicting potential energy transformations and system stability.Key elements include:
  • The First Law: Energy cannot be created or destroyed, only transferred. It is crucial in calculating energy changes (\( \Delta H \)) during reactions.
  • The Second Law: Entropy of an isolated system always increases, dictating the possible spontaneity of processes.
  • The Third Law: As temperature approaches absolute zero, the entropy of a perfect crystal also approaches zero.
Thermodynamics underpins the equations used in concerning Gibbs Free Energy and equilibrium,making it essential for predicting chemical behaviors and processes.

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Most popular questions from this chapter

Hydrogenation reactions (for example, the process of converting \(\mathrm{C}=\mathrm{C}\) bonds to \(\mathrm{C}-\mathrm{C}\) bonds in food industry) are facilitated by the use of a transition metal catalyst, such as \(\mathrm{Ni}\) or \(\mathrm{Pt}\). The initial step is the adsorption, or binding, of hydrogen gas onto the metal surface. Predict the signs of \(\Delta H, \Delta S,\) and \(\Delta G\) when hydrogen gas is adsorbed onto the surface of Ni metal.

State the second law of thermodynamics in words and express it mathematically.

Find the temperatures at which reactions with the following \(\Delta H\) and \(\Delta S\) values would become spontaneous: (a) \(\Delta H=-126 \mathrm{~kJ} / \mathrm{mol}, \Delta S=84 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} ;\) (b) \(\Delta H=-11.7 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-105 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\).

The equilibrium constant \(K_{P}\) for the reaction $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ is \(5.62 \times 10^{35}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{COCl}_{2}\) at \(25^{\circ} \mathrm{C} .\)

A student placed \(1 \mathrm{~g}\) of each of three compounds \(\mathrm{A}\), \(\mathrm{B},\) and \(\mathrm{C}\) in a container and found that after 1 week no change had occurred. Offer some possible explanations for the fact that no reactions took place. Assume that \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are totally miscible liquids.

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