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Chapter 18: Problem 41

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
The value of \( K_{P} \) is approximately 0.322.

Step by step solution

01

Convert 螖G掳 to Joules

First, convert the given standard Gibbs free energy change \( \Delta G^{\circ} = 2.60 \mathrm{~kJ/mol} \) into Joules per mole, since the formula we plan to use requires the energy unit in Joules. Thus, we have \( \Delta G^{\circ} = 2.60 \times 10^{3} \mathrm{~J/mol} \).
02

Use the Relation Between 螖G掳 and KP

Use the relation \( \Delta G^{\circ} = -RT \ln K_{P} \), where \( R = 8.314 \mathrm{~J/(mol \, K)} \), and \( T = 25^{\circ} \mathrm{C} = 298 \mathrm{~K} \) (since the temperature must be in Kelvin). Substitute these values along with \( \Delta G^{\circ} \) to solve for \( K_{P} \).
03

Substitute Known Values into the Formula

Substitute the known values into the equation: \( 2.60 \times 10^{3} = - (8.314)(298) \ln K_{P} \).
04

Solve for ln(KP)

Rearrange the equation to solve for \( \ln K_{P} \): \( \ln K_{P} = \frac{-2.60 \times 10^{3}}{8.314 \times 298} \). Calculate this value.
05

Calculate KP

Use the value of \( \ln K_{P} \) to find \( K_{P} \): \( K_{P} = e^{\ln K_{P}} \). Compute this to get the final result for \( K_{P} \).
06

Final Calculation

After calculating, you should find that \( K_{P} \approx 0.322 \). This completes the calculation based on the given \( \Delta G^{\circ} \).

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Key Concepts

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

Equilibrium Constant
In the study of chemical reactions, the equilibrium constant, represented as \( K_{P} \) for gases, is a fundamental concept. It determines the ratio of the concentrations of products to reactants at equilibrium.
This ratio measures the extent to which a chemical reaction proceeds. A high \( K_{P} \) value indicates products are favored, whereas a low \( K_{P} \) suggests reactants are predominant.
The expression for \( K_{P} \) depends on the balanced chemical equation for the reaction:
  • For the given reaction of \( \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2\mathrm{HI}(g) \), \( K_{P} \) involves the partial pressures of \( \mathrm{HI} \), \( \mathrm{H}_{2} \), and \( \mathrm{I}_{2} \).
  • It's crucial to note the equation is defined at a specific temperature, as \( K_{P} \) is temperature-dependent.
Understanding \( K_{P} \) can help predict how changes in pressure, temperature, or concentration may shift the position of equilibrium.
Thermodynamics
Thermodynamics provides the backdrop for understanding energy changes in reactions, helping us connect these concepts to practical calculations. A primary parameter here is Gibbs Free Energy (
  • \( \Delta G^{\circ} \) indicates if a reaction is spontaneous under constant pressure and temperature. If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous.
The equation \( \Delta G^{\circ} = -RT \ln K_{P} \) links these thermodynamic concepts to equilibrium.
Here:
  • \( R \) is the universal gas constant.
  • \( T \) is the temperature in Kelvin.
This formula lets us calculate \( K_{P} \) from \( \Delta G^{\circ} \), helping to translate energy balance insights into practical predictions about reaction direction.
Reaction Kinetics
While thermodynamics can tell us if a reaction can occur, reaction kinetics explains how fast it occurs. Kinetics focuses on the rate at which reactants turn into products. This is separate from equilibrium since a reaction might be favorable (thermodynamically) but slow (kinetically).
Temperature, pressure, and catalysts can influence rates, but do not affect the equilibrium constant. It's crucial to remember that:
  • While \( \Delta G^{\circ} \) implies a favored direction, kinetics determines the time scale.
  • Fast kinetics doesn't imply high \( K_{P} \); both concepts must work together to fully understand reactions.
Thus, understanding equilibrium and kinetics allows predicting both the possibility and timeline of reactions.
Chemical Equilibrium
Chemical equilibrium represents a state in a reaction where the rates of forward and reverse reactions are equal. This means the concentrations of reactants and products have reached a steady state.
  • At equilibrium, the Gibbs free energy is minimized, making the system very stable.
  • The expression for equilibrium constant \( K_{P} \) only includes gaseous reactants and products.
Shifts in equilibrium, described by Le Chatelier's Principle, illustrate how a system responds to changes in conditions.
  • Increasing pressure or concentration of reactants typically shifts equilibrium towards products.
  • Temperature changes can shift the equilibrium constant \( \left(K_{P}\right) \), influencing reaction spontaneity.
Through understanding chemical equilibrium, one can predict how and why reactions adjust to changing conditions.

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Most popular questions from this chapter

Arrange the following substances ( 1 mole each) in order of increasing entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Ne}(g)\), (b) \(\mathrm{SO}_{2}(g),\) (c) \(\mathrm{Na}(s)\) (d) \(\mathrm{NaCl}(s)\) (e) \(\mathrm{H}_{2}(g)\). Give the reasons for your arrangement.

Consider the following facts: Water freezes spontaneously at \(-5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm},\) and ice has a lower entropy than liquid water. Explain how a spontaneous process can lead to a decrease in entropy.

What is the relationship between entropy and the number of possible arrangements of molecules in a system?

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 \(\Delta G_{i}^{o}\) for \(\mathrm{COCl}_{2}\) at \(25^{\circ} \mathrm{C}\)

Many hydrocarbons exist as structural isomers, which are compounds that have the same molecular formula but different structures. For example, both butane and isobutane have the same molecular formula of \(\mathrm{C}_{4} \mathrm{H}_{10}\) Calculate the mole percent of these molecules in an equilibrium mixture at \(25^{\circ} \mathrm{C}\), given that the standard free energy of formation of butane is \(-15.9 \mathrm{~kJ} / \mathrm{mol}\) and that of isobutane is \(-18.0 \mathrm{~kJ} / \mathrm{mol}\). Does your result support the notion that straight-chain hydrocarbons (i.e. hydrocarbons in which the \(\mathrm{C}\) atoms are joined along a line) are less stable than branch-chain hydrocarbons?
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